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COLLEGE   ALGEBRA 


Br 
HENEY  BURCHARD  PINE 

PaoFESsoR  OF  Mathematics  in  Princeton  University 


GINN  AND  COMPANY 

BOSTON     •    NEW   YORK     •    CHICAGO     •     LONDON 
ATLANTA     •     DALLAS     •    COLUMBUS     •    SAN   FRANCISCO 


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ENTERED   AT    STATIONERS     HALL 

COPYRIGHT,  1901,  1904,  BY 
HENRY  B.  FINE 

ALL  RIGHTS  RESERVED 

PRINTED  IN  THE  UNITED  STATES  OF  AMERICA 

532.10 


'-r 


Wi)t  atbenxam  ]$ttes 


PREFACE 


In  this  book  I  have  endeavored  to  develop  the  theory  of  the 
algebraic  processes  in  as  elementary  and  informal  a  manner 
as  possible,  but  connectedly  and  rigorously,  and  to  present 
the  processes  themselves  in  the  form  best  adapted  to  the 
purposes  of  practical  reckoning. 

The  book  is  meant  to  contain  everything  relating  to  algebra 
that  a  student  is  likely  to  need  during  his  school  and  college 
course,  and  the  effort  has  been  made  to  arrange  this  varied 
material  in  an  order  which  will  properly  exhibit  the  logical 
interdependence  of  its  related  parts. 

It  has  seemed  to  me  best  to  divide  the  book  into  two  parts, 
a  preliminary  part  devoted  to  the  number  system  of  algebra 
and  a  principal  part  devoted  to  algebra  itself. 

I  have  based  my  discussion  of  number  on  the  notion  of 
cardinal  number  and  the  notion  of  order  as  exhibited  in  the 
first  instance  in  the  natural  scale  1,  2,  3,  •••.  There  are  con- 
siderations of  a  theoretical  nature  in  favor  of  this  procedure 
into  which  I  need  not  enter  here.  But  experience  has  con- 
vinced me  that  from  a  pedagogical  point  of  view  also  this 
method  is  the  best.  The  meaning  of  the  ordinal  definition  of 
an  irrational  number,  for  example,  can  be  made  clear  even  to 
a  young  student,  whereas  any  other  real  definition  of  such  a 
number  is  too  abstract  to  be  always  correctly  understood  by 
advanced  students. 

My  discussion  of  number  may  be  thought  unnecessarily 
elaborate.  But  in  dealing  with  questions  of  this  fundamental 
character  a  writer  cannot  with  a  good  conscience  omit  points 
which  properly  belong  to  his  discussion,  or  fail  to  give  proofs 


IV  PREFACE 

of  statements  which  require  demonstration.  I  hope  the  details 
of  the  discussion  will  interest  the  more  thoughtful  class  of 
students ;  but  all  that  the  general  student  need  be  asked  to 
learn  from  it  is  the  ordinal  character  of  the  real  numbers 
and  of  the  relations  of  equality  and  inequality  among  them, 
and  that  for  all  numbers,  real  and  complex,  the  fundamental 
operations  admit  of  definitions  which  conform  to  the  commu- 
tative, associative,  and  distributive  laws. 

In  the  second  or  main  part  of  the  book  I  begin  by  observ- 
ing that  in  algebra,  where  numbers  are  represented  by  letters, 
the  laws  just  mentioned  are  essentially  the  definitions  of  the 
fundamental  operations.  These  algebraic  definitions  are  stated 
in  detail,  and  from  them  the  entire  theory  of  the  algebraic 
processes  and  the  practical  rules  of  reckoning  are  subsequently 
derived  deductively. 

I  shall  not  attempt  to  describe  this  part  of  the  book  mi- 
nutely. It  will  be  found  to  differ  in  essential  features  from 
the  text-books  in  general  use.  I  have  carefully  refrained  from 
departing  from  accepted  methods  merely  for  the  sake  of  nov- 
elty. But  I  have  not  hesitated  to  depart  from  these  methods 
when  this  seemed  to  me  necessary  in  order  to  secure  logical 
consistency,  or  when  I  saw  an  opportunity  to  simplify  a  matter 
of  theory  or  practice.  I  have  given  little  space  to  special 
devices  either  in  the  text  or  in  the  exercises.  On  the  other 
hand,  I  have  constantly  sought  to  assist  the  student  to  really 
mas*ier  the  general  methods  of  the  science. 

Thus,  instead  of  relegating  to  the  latter  part  of  the  book 
the  method  of  undetermined  coefficients,  the  principal  method 
of  investigation  in  analysis,  I  have  introduced  it  very  early 
and  have  subsequently  employed  it  wherever  this  could  be 
done  to  advantage.  This  has  naturally  affected  the  arrange- 
ment of  topics.  In  particular  I  have  considered  partial  frac- 
tions in  the  chapter  on  fractions.  They  belong  there  logically, 
and  when  adequately  treated,  supply  the  best  practice  in  ele- 
mentary reckoning  that  algebra  affords. 


PREFACE  V 

Again,  I  have  laid  great  stress  upon  the  division  transfor- 
mation and  its  consequences,  and  in  connection  with  it  have 
introduced  the  powerful  method  of  synthetic  division. 

The  earlier  chapters  on  equations  will  be  found  to  contain  a 
pretty  full  discussion  of  the  reasoning  on  which  the  solution  of 
equations  depends,  a  more  systematic  treatment  than  is  custom- 
ary of  systems  of  equations  which  can  be  solved  by  aid  of  the 
quadratic,  and  a  somewhat  elaborate  consideration  of  the  graphs 
of  equations  of  the  first  and  second  degrees  in  two  variables. 

The  binomial  theorem  for  positive  integral  exponents  is 
treated  as  a  special  case  of  continued  multiplication,  experience 
having  convinced  me  that  no  other  method  serves  so  well  to 
convey  to  the  student  the  meaning  of  this  important  theorem. 
I  have  introduced  practice  in  the  use  of  the  general  binomial 
theorem  in  the  chapter  on  fractional  exponents,  but  have 
deferred  the  proof  of  the  theorem  itself,  together  with  all 
that  relates  to  the  subject  of  infinite  series,  until  near  the  end 
of  the  book. 

In  the  chapters  on  the  theory  of  equations  and  determinants 
there  will  be  found  proofs  of  the  fundamental  theorems  regard- 
ing symmetric  functions  of  the  roots  of  an  equation  and  a 
discussion  of  the  more  important  properties  of  resultants. 
These  subjects  do  not  belong  in  an  elementary  course  in 
algebra,  but  the  college  student  who  continues  his  mathemat- 
ical studies  will  need  them.  The  like  is  to  be  said  of  the  chap- 
•  ters  on  infinite  series  and  of  the  chapter  on  properties  of 
continuous  functions  with  which  the  book  ends. 

The  ideas  which  underlie  the  first  part  of  the  book  are 
those  of  Rowan  Hamilton,  Grassmann,  Helmholtz,  Dedekind, 
and  Georg  Cantor.  But  I  do  not  know  that  any  one  hitherto 
has  developed  the  doctrine  of  ordinal  number  from  just  the 
point  of  view  I  have  taken,  and  in  the  same  detail. 

In  preparing  the  algebra  itself  I  have  profited  by  sugges- 
tions from  many  books  on  the  subject.  I  wish  in  particular 
to  acknowledge  my  indebtedness  to  the  treatises  of  Chrystal. 


vi  PREFACE 

The  book  has  been  several  years  in  preparation.  Every 
year  since  1898  the  publishers  have  done  me  the  courtesy  to 
issue  for  the  use  of  the  freshmen  at  Princeton  a  pamphlet 
containing  what  at  the  time  seemed  to  me  the  most  satisfac- 
tory treatment  of  the  more  important  parts  of  algebra.  With 
the  assistance  of  my  colleagues,  Mr.  Eisenhart  and  Mr.  Gillespie, 
I  endeavored  after  each  new  trial  to  select  what  had  proved 
good  and  to  discard  what  had  proved  unsatisfactory.  As  a 
consequence,  much  of  the  book  has  been  rewritten  a  number 
of  times.  No  doubt  subsequent  experience  will  bring  to  light 
many  further  possibilities  of  improvement ;  but  I  have  hopes 
that  as  the  book  stands  it  will  serve  to  show  that  algebra  is 
not  only  more  intelligible  to  the  student,  but  also  more  inter- 
esting and  stimulating,  when  due  consideration  is  given  to  the 
reasoning  on  which  its  processes  depend. 

HENRY  E.  FINE 


Princeton  University 
June,  1905 


K' 


CONTENTS 


PAET  FIRST  —  NUMBERS 

PAGE 

I.    The  Natural  Numbers  —  Counting,  Addition,  and  Mul- 
tiplication      1 

II.    Subtraction  and  the  Negative 16 

III.  Division  and  Fractions 27 

IV.  Irrational  Numbers S9 

V.    Imaginary  and  Complex  Numbers 70 


PART   SECOND  —  ALGEBRA 

I.  Preliminary  Considerations 79 

II.  The  Fundamental  Operations  .  > 93 

III.  Simple  Equations  in  one  Unknown  Letter 110— 

IV.  Systems  of  Simultaneous  Simple  Equations      ....  127—— 
V.  The  Division  Transformation 155-— 

VI.  Factors  of  Rational  Integral  Expressions       ....   176 

VII.  Highest  Common  Factor  and  Lowest  Common  Multiple  196 

V^III.  Rational  Fractions 213 

IX.  Symmetric  Functions 245 

X.  The  Binomial  Theorem 252— _ 

XI.  Evolution 260 

XII.  Irrational       Functions.      Radicals      and       Fractional 

Exponents 271 

XIII.  Quadratic  Equations 298——' 

XIV.  Discussion   of   the  Quadratic    Equation.      Maxima  and 

Minima 304..^ 

XV.    Equations  of  Higher  Degrek  which  can  be   solved  by 

Means  of  Quadratics 309.- 

vii 


vm  CONTENTS 

PAGE 

XVI.    Simultaneous  Equations  which   can  be   solved   by 

Means  of  Quadratics 317 

XVII.    Inequalities 340 

XVIII.    Indeterminate  Equations  of  the  First  Degree       .  342 

XIX.    Ratio  and  Proportion.     Variation 347 

XX.    Arithmetical  Progression 3.54 

XXI.    Geometrical  Progression 357 

XXII.    Harmonical  Progression 362 

XXIII.    Method    of   Differences.     Arithmetical    Progres- 
sions OF  Higher  Orders.    Interpolation      .     .     .  364 

.  XXIV.    Logarithms 374 

XXV.    Permutations  and  Combinations 393 

XXVI.    The  Multinomial  Theorem 408 

XXVII.    Probability 409 

XXVIII.    Mathematical  Induction 424 

XXIX.    Theory  of  Equations 425 

XXX.    Cubic  and  Biquadratic  Equations 483 

XXXI.    Determinants  and  Elimination 492 

XXXII.    Convergence  of  Infinite  Series     .     .     .     .    y  .     .   520 

XXXIII.  Operations  with  Infinite  Series 539 

XXXIV.  The  Binomial,  Exponential,  and  Logarithmic  Series  553 
XXXV.    Recurring  Series 560 

XXXVI.    Infinite  Products 564 

XXXVII.    Continued  Fractions 666 

XXXVIII.    Properties  of  Continuous  Functions 577 

INDEX 591 


A  COLLEGE  ALGEBRA 

PART   FIRST— NUMBERS 


I.     THE   NATURAL   NUMBERS  — COUNTING, 
ADDITION,  AND    MULTIPLICATION 

GROUPS   OF  THINGS  AND  THEIR   CARDINAL  NUMBERS 

Groups  of  things.  In  our  daily  experience  things  present 
themselves  to  our  attention  not  only  singly  but  associated  in 
groups  or  assemblages. 

The  fingers  of  a  hand,  a  herd  of  cattle,  the  angular  points 
of  a  polygon  are  examples  of  such  groups  of  things. 

We  think  of  certain  things  as  constituting  a  group,  when 
we  distinguish  them  from  other  things  not  individually  but 
as  a  whole,  and  so  make  them  collectively  a  single  object  of 
our  attention. 

For  convenience,  let  us  call  the  things  which  constitute  a 
group  the  eleynents  of  the  group. 

Equivalent  groups.  One-to-one  correspondence.  The  two  groups 
of  letters  ABC  and  DEF  are  so  related  that  we  can  combine 
all  their  elements  in  pairs  by  matching  elements  of  the  one 
with  elements  of  the  other,  one  element  with  one  element. 
Thus,  we  may  match  A  with  D,  B  with  E,  and  C  with  F. 

Whenever  it  is  possible  to  match  all  the  elements  of  two 
groups  in  this  manner,  we  shall  say  that  the  groups  are 
equivalent;  and  the  process  of  matching  elements  we  shall 
call  bringing  the  groups  into  a  one-to-one  relation,  or  a  relation 
of  one-to-one  correspondence. 

1 


2  \  A  \  (JOjiXtGE    ALGEBRA 

TIxGOi e>n.,  ;,^_  (wa  ^/y;/^,^.^ (^rc  equivalent  to  the  same  third 
group,  they  are  equivalent  to  one  another. 

For,  by  hypothesis,  we  can  bring  each  of  the  two  groups 
into  one-to-one  correspondence  with  the  third  group.  But 
the  two  groups  will  then  be  in  one-to-one  correspondence 
with  each  other,  if  we  regard  as  mates  every  two  of  their 
elements  which  we  have  matched  with  the  same  element  of 
the  third  group. 

Cardinal  number.  We  may  think  of  all  possible  groups  of 
things  as  distributed  into  classes  of  equivalent  groups,  any 
two  given  groups  belonging  to  the  same  class  or  to  different 
classes,  according  as  it  is,  or  is  not,  possible  to  bring  them 
into  one-to-one  correspondence. 

Thus,  the  groups  of  letters  ABCD  and  EFGH  belong  to  the 
same  class,  the  groups  ABCD  and  EFG  to  different  classes. 

The  property  which  is  common  to  all  groups  of  one  class, 
and  which  distinguishes  the  groups  cf  one  class  from  those  of 
another  class,  is  the  number  of  things  in  a  group,  or  its  cardinal 
number.     In  other  words, 

The  number  of  things  in  a  group,  or  its  cardinal  number,  is 
that  property  which  is  common  to  the  group  itself  and  every 
group  which  may  he  brought  into  one-to-one  correspondence 
with  it. 

Or  we  may  say  :  "  The  cardinal  number  of  a  group  of  things 
is  that  property  of  the  group  which  remains  unchanged  if  we 
rearrange  the  things  within  the  group,  or  replace  them  one 
by  one  by  other  things "  ;  or  again,  "  it  is  that  property  of 
the  group  which  is  independent  of  the  character  of  the  things 
themselves  and  of  their  arrangement  within  the  group." 

For  rearranging  the  things  or  replacing  them,  one  by  one, 
by  other  things  will  merely  transform  the  group  into  an  equiva- 
lent group,  §  2.  And  a  property  which  remains  unchanged 
during  all  such  changes  in  the  group  must  be  independent  of 
the  character  of  the  things  and  of  their  arrangement. 


THE    NATURAL    NUMBERS  3 

Part.  We  say  that  a  first  group  is  a  part  of  a  second  group 
■when  the  elements  of  the  first  are  some,  but  not  all,  of  the 
elements  of  the  second. 

Thus,  the  group  ABC  is  a  part  of  the  group  ABCD. 

From  this  definition  it  immediately  follows  that 

If  the  first  of  three  gt'oups  be  a  jjart  of  the  second,  arid  the 
second  a  j^^f^'t  of  t/ie  third,  then  the  first  is  also  a  part  of 
the  third. 

_glnite  and  infinite  groups, Wa,-sa,y-that  a  group  or  assem- 
blage is  finite  when  it  is  equivalent  to  no  one  of  its  parts ; 
infinite  when  it  is  equivalent  to  certain  of  its  parts.* 

Thus,  the  group  ABC  is  finite;  for  it  cannot  be  brought  into  one-to- 
one  correspondence  with  BC,  or  with  any  other  of  its  parts. 

But  any  never-ending  sequence  of  marks  or  symbols,  the  never-ending 
sequence  of  numerals  1,  2,  3,  4,  •  •  • ,  for  example,  is  an  infinite  assemblage. 

We  can,  for  instance,  set  up  a  one-to-one  relation  between  the  entire 
assemblage  1,  2,  3,  4,  •  •  •  and  that  part  of  it  which  begins  at  2,  namely, 

between  1,  2,  3,  4,  5,  •  •  •  (a) 

and  2,3,4,5,6,...,  (b) 

by  matching  1  in  (a)  with  2  in  (b),  2  in  (a)  with  3  in  (b),  and  so  on,  — there 
being  for  every  numeral  that  we  may  choose  to  name  in  (a)  a  corresponding 
numeral  in  (b). 

Hence  the  assemblage  (a)  is  equivalent  to  its  part  (b).  Therefore  (a) 
is  infinite. 

Less  and  greater  cardinal  numbers.     Let  M  and  iV  denote  any     8 
two  finite  groups.     It  must  be  the  case  that 

1.    M  and  N  are  equivalent, 
or  2.    M  is  equivalent  to  a  part  of  N, 

or  3.    N  is  equivalent  to  a  part  of  M. 

*  Of  course  we  cannot  actually  take  account  of  all  the  elements,  one  by  one, 
of  an  infinite  group  —  or  assemblage,  as  it  is  more  often  called.  We  regard 
such  an  assemblage  as  defined  when  a  law  has  been  stated  which  enables  us 
to  say  of  every  given  thing  whether  it  belongs  to  the  assemblage  or  not. 


4  A   COLLEGE   ALGEBRA 

In  the  first  case  we  say  that  M  and  N  have  the  same  cardinal 
number,  §  4,  or  equal  cardinal  numbers  ;  in  the  second  case,  that 
the  cardinal  number  of  M  is  less  than  that  oi  N ;  in  the  third, 
that  the  cardinal  number  of  M  is  greater  than  that  of  N. 

Thus,  if  M  is  the  group  of  letters  c6c,  and  N  the  group  de/gr,  then  M 
is  equivalent  to  a  part  of  N,  to  the  part  de/,  for  example. 

Hence  the  cardinal  number  of  M  is  less  than  that  of  N,  and  the  cardinal 
number  of  N  is  greater  than  that  of  M. 

9  Note.     It  follows  from  the  definition  of  finite  group,  §  7,  that  there  is 

no  ambiguity  about  the  relations  "equal,"   "greater,"  and  "less"  as 
here  defined. 

Thus,  the  definition  does  not  make  it  possible  for  the  cardinal  number 
of  M  to  be  at  the  same  time  equal  to  and  less  than  that  of  N,  since  this 
would  mean  that  M  is  equivalent  to  N  and  also  to  a  part  of  N,  therefore 
that  N  is  equivalent  to  one  of  its  parts,  §  3,  and  therefore,  finally,  that  N 
is  infinite,  §  7. 

10  Corollary.  If  the  first  of  three  cardinal  numbers  be  less  than 
the  second,  and  the  second  less  than  the  third,  then  the  first  is 
also  less  than  the  third. 

For  if  Jf,  N,  P  denote  any  groujis  of  things  of  which  these  are  the 
cardinals,  M  is  equivalent  to  a  part  of  N,  and  JV  to  a  part  of  P;  therefore 
M  is  equivalent  to  a  part  of  P,  §§  3,  G. 

11  The  system  of  cardinal  numbers.  By  starting  with  a  group 
which  contains  but  a  single  element  and  repeatedly  "  adding  " 
one  new  thing,  we  are  led  to  the  following  list  of  the  cardinal 
numbers : 

1.  The  cardinal  number  of  a  "  group  "  like  I,  which  contains 
but  a  single  element. 

2.  The  cardinal  number  of  a  group  like  II,  obtained  by  adding 
a  single  element  to  a  group  of  the  first  kind. 

3.  The  cardinal  number  of  a  group  like  III,  obtained  by  adding 
a  single  element  to  a  group  of  the  second  kind. 

4.  And  so  on,  without  end. 

We  name  these  successive  cardinals  "one,"  " two,"*' three,"  •••, 
and  represent  them  by  the  signs  1,  2,  3,  •  •  • . 


THE   NATURAL   NUMBERS       '  5 

Observations  on  this  system.     Calling  the  cardinal  number     12 
of  any  finite  group  a  finite  cardinal,  we  make  the  following 
observations  regarding  the  list  of  cardinals  which  has  just 
been  described. 

First.     Every  cardinal  contained  in  this  list  is  finite. 

For  the  group  I  is  finite,  since  it  has  no  part  to  which  to  be  equiva^  " 
lent,  §  7 ;  and  each  subsequent  group  is  finite,  because  a  group  obtained 
by  adding  a  single  thing  to  a  finite  group  is  itself  finite.*    Thus,  II  is  finite 
because  I  is ;  III  is  finite  because  II  is  ;  and  so  on. 

Second.     Every  finite  cardinal  is  contained  in  the  list. 

For,  by  definition,  every  finite  cardinal  is  the  cardinal  number  of  some 
finite  group,  as  M.  But  we  can  construct  a  group  of  marks  III  •  •  •  I  equiva- 
lent to  any  given  finite  group  if,  by  making  one  mark  for  each  object  in 
M.  And  this  group  of  marks  must  have  a  last  mark,  and  therefore  be 
included  in  the  list  of  §  11,  since  otherwise  it  would  be  never-ending  and 
therefore  itself,  and  with  it  M,  be  infinite,  §  7. 

Third.     No  two  of  these  cardinals  are  equal. 

This  follows  from  the  definition  in  §  8.  For,  as  just  shown,  all  of  the 
groups  I,  II,  III,  •  •  •  are  finite  ;  and  it  is  true  of  every  two  of  them  that  one 
is  equivalent  to  a  part  of  the  other. 

*  We  may  prove  this  as  follows  (G.  Cantor,  Math.  Ann.,  Vol.  46,  p.  490) : 
If  M  denote  a  finite  group,  and  e  a  single  thing,  the  group  Me,  obtained 
by  adding  e  to  M,  is  also  finite. 

For  let  G  =  H  denote  that  the  groups  G  and  H  are  equiA'alent. 

If  Me  is  not  finite,  it  must  be  equi^^^lent  to  some  one  of  its  parts,  §  7. 

Let  P  denote  this  part,  so  that  Me  =  P. 

(1)  Suppose  that  P  does  not  contain  e. 

Let/  denote  the  element  of  P  which  is  matched  with  e  in  Me,  and  represent 
the  rest  of  P  by  Pi.  _ 

Then  sinoe  Mp  =  P\  f  and  e  =  /",  we  have  M=  P\. 
But  tliis  is  inipossibie,  since  M  is  finite  and  Pi  is  a  part  of  M,  §  7. 

(2)  Sujipfisc  tliat  P  does  contain  e. 

It  cannot  be  tliat  e  in  P  is  matched  with  e  in  Me,  for  then  the  rest  of  P, 
which  is  a  part  of  M,  would  be  equivalent  to  M. 

But  suppose  that  e  in  P  is  matched  with  some  other  element,  as  g  in  Me,  and 
that  e  in  Me  is  matched  with/ in  P. 

If  Me  =  P  oe  true  on  this  hypothesis,  it  must  also  be  true  if  we  recombine 
the  elements  e,  /,  g  so  as  to  match  r  in  P  with  e  in  Me,  and  /  in  P  with  g 
in  Me.  But,  as  "ins't  shown,  we  should  then  have  a  part  of  P  equivalent  to  M. 
Hence  this  hypothesis  albo  is  impossible 


A   COLLEGE   ALGEBRA 


THE  NATURAL  SCALE.     EQUATIONS  AND  INEQUALITIES 

13  The  natural  numbers.     We  call  the  signs  1,  2,  3, oi 

their  names  "  one,"  "  two,"  "  three," positive  integers  or 

natural  numbers.     Hence 

A  natural  number  is  a  sigyi  or  symbol  for  a  cardinal  number. 

14  The  natural  scale.  Arranging  these  numbers  in  an  order 
corresponding  to  that  already  given  the  cardinals  which  they 
represent,  §  11,  we  have  the  never-ending  sequence  of  signs 

1,2,3,4,  5,    ••, 

or  "one,"  "two,"  "three,"  "four,"  "five,"  •••,  which  we  call 
the  natural  scale,  or  the  scale  of  the  natural  numbers. 

15  Each  sign  i?i  the  scale  indicates  the  number  of  the  signs  in 
that  part  of  the  scale  which  it  terminates. 

Thus,  4  indicates  the  number  of  the  signs  1,  2,  3,  4.  For  the  number 
of  signs  1,  2,  .3,  4  is  the  same  as  the  number  of  groups  I,  II,  III,  llli,  and  this, 
in  turn,  is  the  same  as  the  number  of  marks  in  the  last  group,  llll,  §  8. 
And  so  in  general. 

16  The  ordinal  character  of  the  scale.  The  natural  scale,  by  itself 
considered,  is  merely  an  assemblage  of  different  signs  in  which 
there  is  a  first  sign,  namely  1 ;  to  this  a  definite  next  follow- 
ing sign,  namely  2 ;  to  this,  in  turn,  a  definite  next  following 
sign,  namely  3 ;  and  so  on  without  end. 

In  other  words,  the  natural  scale  is  merely  an  assemblage 
of  different  signs  which  follow  one  another  in  a  definite  and 
knoivn  order,  and  having  a  first  but  no  last  sign. 

Regarded  from  this  point  of  view,  the  natural  numbers  themselves  are 
merely  marks  of  order,  namely  of  the  order  in  which  they  occur  —  with 
respect  to  time  —  when  the  scale  is  recited. 

17  It  is  evident  that  the  scale,  in  common  with  all  other  assem- 
blages whose  elements  as  given  us  are  arranged  in  a  definite , 
and  known  order,  has  the  following  properties : 


THE   NATURAL   NUMBERS  7 

1.  We  may  say  of  any  two  of  its  elements  that  the  one 
"precedes"  and  the  other  "follows,"  and  these  words  "pre- 
cede "  and  "  follow  "  have  the  same  meaning  when  applied  to 
any  one  pair  of  the  elements  as  when  applied  to  any  other  pair. 

2.  If  any  two  of  the  elements  be  given,  we  can  always  deter- 
mine ivhich  precedes  and  which  follows. 

3.  If  a,  b,  and  c  denote  any  three  of  the  elements  such  that 
a  precedes  b,  and  b  precedes  c,  then  a  precedes  c. 

An  assemblage  may  already  possess  these  properties  when 
given  us,  or  we  may  have  imposed  them  on  it  by  some  rule  of 
arrangement  of  our  own  choosing.  In  either  ease  we  call  the 
assemblage  an  ordinal  system. 

Instances  of  the  first  kind  are  (1)  the  natural  scale  itself ;  (2)  a  sequence 
of  events  in  time ;  (3)  a  row  of  points  ranged  from  left  to  right  along 
a  horizontal  line.  An  instance  of  the  second  kind  is  a  group  of  men 
arranged  according  to  the  alphabetic  order  of  their  names. 

An  assemblage  may  also  have  "  coincident "  elements.     Thus,     18 
in  a  group  of  events  two  or  more  may  be  simultaneous. 

We  call  such  an  assemblage  ordinal  when  the  relations  1, 2, 3 
hold  good  among  its  7io7i-co incident  elements  —  it  being  true  of 
the  coincident  elements  that 

4.  If  a  coincides  with  b,  and  b  with  c,  then  a  coincides 
with  c. 

5.  If  a  coincides  with  b,  and  b  precedes  c,  then  a  precedes  c. 

It  is  by  their  relative  order  in  the  scale  that  the  natural     19 
numbers  indicate  relations   of  greater   and  less   among  the 
cardinal  numbers. 

For  of  any  two  given  cardinals  that  one  is  greater  whose 
natural  number  occurs  later  in  the  scale. 

And  the  relation :  "  if  the  first  of  three  cardinals  be  less 
than  the  second,  and  the  second  less  than  the  third,  then 
the  first  is  less  than  the  third,"  is  represented  in  the  scale 
by  the  relation :  "  if  a  precede  b,  and  b  precede  c,  then  a 
precedes  c." 


8  A   COLLEGE    ALGEBRA 

In  fact,  we  seldom  employ  any  other  method  than  this  for  comparing 
cardinals.  We  do  not  compare  the  cardinal  numbers  of  groups  of  things 
directly,  by  the  method  of  §  8.  On  the  contrary,  we  represent  them  by  the 
appropriate  natural  numbers,  and  infer  which  are  greater  and  which  less 
from  the  relative  order  in  which  these  natural  numbers  occur  in  the  scale. 
The  process  causes  us  no  conscious  effort  of  thought,  for  the  scale  is  so 
vividly  impressed  on  our  minds  that,  when  any  two  of  the  natural  numbers 
are  mentioned,  we  instantly  recognize  which  precedes  and  which  follows. 
Thus,  if  we  are  told  of  two  cities,  A  and  B,  that  the  population  of  A  is 
120,000,  and  that  of  B,  125,000,  we  immediately  conclude  that  B  has  the 
greater  number  of  inhabitants,  because  we  know  that  125,000  occurs  later 
in  the  scale  than  120,000  does. 

20  Numerical  equations  and  inequalities.  In  what  follows,  the 
word  "  number "  will  mean  natural  number,  §  13 ;  and  the 
letters  a,  h,  c  will  denote  any  such  numbers. 

21  When  we  wish  to  indicate  that  a  and  b  denote  the  same  num- 
ber, or  "  coincide  "  in  the  natural  scale,  we  employ  the  equation 

a  =  b,  read  "  a  equals  b." 

22  But  when  we  wish  to  indicate  that  a  precedes  and  J  follows 
in  the  natural  scale,  we  employ  one  of  the  inequalities 

a  <b,  read  "  a  is  less  than  b  "  ; 
b  >  a,  read  "  b  is  greater  than  a." 

23  Of  course,  strictly  speaking,  these  words,  "equal,"  "less," 
and  "  greater,"  refer  not  to  the  signs  a  and  b  themselves,  but 
to  the  cardinals  which  they  represent.  Thus,  the  phrase,  "  a 
is  less  than  b,"  is  merely  an  abbreviation  for,  "the  cardinal 
which  a  represents  is  less  than  the  cardinal  which  b  represents." 

But  all  that  the  inequality  a  <b  means /or  the  signs  a  and 
b  themsdoes  is  that  a  precedes  b  in  the  scale. 

24  Rules  of  equality  and  inequality.  From  §§  17,  18  and  these 
definitions,  §§21,  22,  it  immediately  follows  that 

1.  If  a  =  Z' and  Z»  =  c,  then  a  =  c. 

2.  \i  a<b  and  b  <c,  then  a<c. 

3.  li  a  =  b  and  b  <  c,  tnen  a  <c. 


THE   NATURAL   NUMBERS 


COUNTING 


Arithmetic  is  primarily  concerned  with  the  ordinal  relations     25 
existing  among  the  natural  numbers,  and  with  certain  opera- 
tions by  which  these  numbers  may  be  combined. 

The  operations  of  arithmetic  have  their  origin  in  counting. 

Counting.     To  discover  what  the  cardinal  number  of  a  given     26 
group  of  objects  is,  we  count  the  group. 

The  process  is  a  very  familiar  one.  We  label  one  of  the 
objects  "  one,"  another  <'  two,"  and  so  on,  until  there  are  no 
objects  left  —  being  careful  to  use  these  verbal  signs  ''one," 
"two,"  •  •  •,  without  omissions,  in  the  order  of  their  occurrence 
in  the  scale,  but  selecting  the  objects  themselves  in  any  order 
that  suits  our  whim  or  convenience  ;  and  the  sign  or  label  with 
which  the  process  ends  is  what  we  seek,  —  the  name  of  the 
cardinal  number  of  the  group  itself.  For  owing  to  the  ordinal 
character  of  the  scale,  this  last  sign  indicates  how  many  signs 
have  been  used  all  told,  §  15,  and  therefore  how  many  objects 
there  are  in  the  group,  §  8. 

Thus,  the  process  of  counting  may  be  described  as  bringing 
the  group  counted  into  one-to-one  correspondence,  §  2,  with  a 
part  of  the  natural  scale  —  namely,  the  part  which  begins  at 
"  one  "  and  ends  with  the  last  number  used  in  the  count. 

Observe  that  the  natural  numbers  serve  a  double  purpose  in 
counting  :  (1)  We  use  a  certain  group  of  them  as  mere  counters 
in  carrying  out  the  process,  and  (2)  we  employ  the  last  one  so 
used  to  record  the  result  of  the  count. 

We  have  intimated  that  it  is  immaterial  in  what  order  we 
select  the  objects  themselves.     This  may  be  proved  as  follows : 

Theorem.      The  result  of  counting  a  finite  group  of  objects  is     27 
the  same,  xuhatever  the  order  in  which  we  select  the  objects. 

Suppose,  for  example,  that  the  result  of  counting  a  certain 
group  were  99  when  the  objects  are  selected  in  one  order,  P, 
but  97  when  they  are  selected  in  another  order,  Q. 


10  A   COLLEGE    ALGEBRA 

The  group  which  consists  of  the  first  97  objects  in  the  ordeT 
P  would  then  be  equivalent  to  the  entire  group  in  the  order  Q, 
for,  by  hypothesis,  both  have  been  matched  with  the  first  97 
numbers  of  the  natural  scale,  §  3. 

But  this  is  impossible,  since  it  would  make  a  part  of 
the  group  equivalent  to  the  whole ;  whereas  the  group  is,  by 
hypothesis,  a  finite  group,  §  7. 

28  Another  definition  of  cardinal  number.  We  may  make  the 
theorem  just  demonstrated  the  basis  of  a  definition  of  the 
cardinal  number  of  a  finite  group,  namely  : 

The  cardinal  nnmber  of  a  finite  group  of  things  is  that  p>rop- 
erty  of  the  group  because  of  which  we  shall  arrive  at  the  same 
natural  number  in  whatever  order  we  count  the  group. 

This  is  the  definition  of  cardinal  number  to  which  we  are  naturally 
led  if  we  choose  to  make  the  natural  scale,  defined  as  in  §  16,  our  starting 
point  in  the  discussion  of  number. 

ADDITION 

29  Definition  of  addition.  To  add  3  to  5  is  to  find  what  number 
occupies  the  third  place  after  5  in  the  natural  scale. 

We  may  find  this  number,  8,  by  counting  three  numbers 
forward  in  the  scale,  beginning  at  6,  thus  :  6,  7,  8. 

We  indicate  the  operation  by  the  sign  +,  read  ''plus," 
writing  5  +  3  =  8. 

And  in  general,  to  add  b  to  a  is  to  find  what  number  occupies 
the  ^th  place  after  a  in  the  natural  scale. 

Since  there  is  no  last  sign  in  the  scale,  this  number  may 
always  be  found.  We  call  it  the  sum  of  a  and  b  and  represent 
it  in  terms  of  a  and  b  by  the  expression  a  -\-  b. 

30  Note.  The  process  of  finding  a  +  b  hy  counting  forward  in  the  scale 
corresponds  step  for  step  to  that  of  adding  to  a  group  of  a  things  the 
elements  of  a  group  of  b  things,  one  at  a  time.  Hence  (1)  the  result  of 
the  latter  process  is  a  group  oi  a  +  b  things,  §  8,  and  (2)  if  a  and  b  denote 
finite  cardinals,  so  also  does  a  +  b.     See  footnote,  p.  5. 


THE    NATURAL    NUMBERS  11 

Since  a  +  1,  a  +  2,  and  so  on,  denote  the  1st,  2d,  and  so     31 
on,  numbers  after  a,  the  sequence  a  +  1,  a  -\-  2,  •  •  •  denotes  all 
that  portion  of  the  scale  which  follows  a. 

Hence  any  given  number  after  a  may  be  expressed  in  the 
form  a  +  d,  where  d  denotes  a  definite  natural  number. 

The  process.     To  add  large  numbers  by  counting  would  be     32 
very  laborious.     We  therefore  memorize  sums  of  the  smaller 
numbers  (addition  tables)  and  from  these  derive  sums  of  the 
larger  numbers  by  applying  the  so-called  "laws"  of  addition 
explained  in  the  following  sections. 

The  laws  of  addition.     Addition  is  a  "  commutative  "  and  an     33 
"  associative  "  operation ;  that  is,  it  conforms  to  the  following 
two  laws  : 

The  commutative  law.     a  -{-  b  —  b  -{-  a,  34 

The  result  of  adding  h  to  2i  is  the  same  as  that  of  adding 
a  to  b. 

The  associative  law.     a  +  (1)  -\-  c)  =  (a  -\-  b)  +  c,  35 

The  result  of  first  adding  c  ^o  b  amd  then  adding  the  sum  so 

obtained  to  a,  is  the  same  as  that  of  first  adding  b  ^o  a  and 

then  adding  c  to  the  sum  so  obtained. 

Note.     In  practice,  we  replace  the  expression  (a  -)-  6)  +  c  by  a  +  6  +  c,     36 
our  understanding  being  tliat  the  expression  a  +  6  +  c  +  •  •  ■  represents 
the  result  of  adding  6  to  a,  c  to  the  sum  so  obtained,  and  so  on. 

Proofs  of  these  laws.     We  may  prove  these  laws  as  follows.       37 

Pirst.     The  co7nmutative  law :  a  -{-  b  =  b  -{-  a. 

Thus,  the  sums  3  +  2  and  2  +  3  are  equal. 

For  3  +  2  represents  the  number  found  by  first  counting  off  three 
numbers,  and  after  that  two  numbers,  on  the  natural  scale.     Thus, 

The  group  counted  1,  2,  3,     4,  5,  (a) 

the  counters,  1,  2,  3,     1,  2.  (b) 

But  as  there  is  a  one-to-one  relation  between  the  groups  of  signs  (a)  and 
(b),  and  every  one-to-one  relation  is  reciprocal,  §  2,  we  may  interchange 
the  roles  of  (a)  and  (b)  ;  that  is,  if  we  make  (b)  the  group  counted,  (a)  will 
represent  the  group  of  counters. 


12  A   COLLEGE    ALGEBRA 

Hence  finding  3  +  2  is  equivalent  to  counting  the  group  of  signs 

1,  2,  3,     1,  2.  (b) 

In  like  manner,  finding  2  +  3  is  equivalent  to  counting  the  group 

1,  2,     1,  2,  3.  (c) 

But  as  (b)  and  (c)  consist  of  the  same  signs  and  differ  only  in  the 

manner  in  w^hich  these  signs  are  arranged,  the  results  of  counting  them 

are  the  same,  §  27  ;  that  is, 

3  +  2  =  2  +  3. 

Similarly  for  any  two  natural  numbers,  a  and  h. 

Second.     The  associative  laiv  :  a  -\-(h  -\-  c)  =  (a  -\-  b)  -j-  c. 

For  in  counting  to  tlie  6th  sign  after  a,  namely  to  a  +  b,  and  then  to 
the  cth  sign  after  this,  namely  to  (a  +  6)  +  c,  we  count  6  +  c  signs  all  told, 
and  hence  arrive  at  the  (6  +  c)th  sign  after  a,  namely  at  a  +  (6  +  c). 

The  notion  of  cardinal  number  is  involved  in  the  proofs  just 
given.  But  addition  may  be  defined  and  its  laws  established 
independently  of  this  notion,  as  is  shown  in  the  footnote  below.* 

*  The  Italian  mathematician  Peano  has  defined  the  system  of  natural  num- 
bers without  using  the  notion  of  cardinal  number,  by  a  set  of  "postulates" 
which  we  may  state  as  follows —  where  "  number  "  means  "  natural  number." 

1.  The  sign  1  is  a  number. 

2.  To  each  number  a  there  is  a  next  following  number  —  call  it  a  +. 

3.  This  number  a  +  is  never  1.  4.    If  a +=  6  + ,  then  a=  6. 

5.   Every  given  number  a  is  present  in  the  sequence  1, 1  +,  (1  +)+,••• . 
The  numerals  2,  3,  •  •  •  are  defined  thus :  2=l  +  ,3=2  +  ,---. 
The  sum  a+6  is  to  mean  the  number  determined  (because  of  5)  by  the 
series  of  formulas  a  +  1  =  a  +,  a  +  2=  (o  +  1)  +,  •  •  • . 

The  series  of  formulas  just  written  is  equivalent  to  the  single  formula 

6.    a  +  (6  +  l)=  (a  +  b)  +  l. 
From  6,  by  "  mathematical  induction,"  we  may  deriA'e  the  laws  of  addition: 

7.   a  +  (b  +  c)=  (a+b) +c.      8.   a  +  b=b  +  a. 
First.    If  7  is  true  when  c=  ^,  it  is  also  true  when  c=  A;  +  1.     For,  by  6  and  7, 
a  +  [b  +  {k  +  l)]--a  +  [{b  +  k)  +  l]=[a  +  {b  +  k)]  +  l 

=  [{a  +  b)  +  k]  +  l=  {a+b)+(k  +  l). 
But,  by  6,  7  is  true  when  c=  1. 

Hence  7  is  true  when  c=  2,  .-.  when  c=  3,  .•.•■•  when  c=  any  number,  by  5. 
Second.    We  first  prove  8  for  tlie  particular  case :  8'.     a  +  1  =  1  +  a. 
If  8' is  true  for  a=  k,  then  {k  +  1)  +  \=  (1  +  k)  +  1=  1  +  (k+  1),  by  G. 
Hence  if  8'  is  true  for  a=  k,  it  is  also  true  for  a=  k+1. 
Hence,  since 8'  is  true  fora=  1,  it  is  true  fora-^  2,  .-.  for  a=  3,  •  ■  •. 
Finally,  if  8  be  true  for  b=k,  it  is  true  for  b=k  +  l.    For,  by  7  and  8', 
a  +  (k  +  1)  =  {a  +  k)  + 1=  \  +  (a  +  k) 

=  l  +  (k  +  a)=(l  +  k)+  a={k  +  l)  +  a. 
Hence,  since  8  is  true  (by  8')  when  6  =  1,  it  is  true  for  6=2,  .-.  for  6  =  3,  .-.  ■  •  • . 
See  Stolz  and  Gmeiner,  Theoret'iarhc  Arithtnrtik,  pp.  13  ff.,  and  tlio  refer- 
ences to  Peano  there  given;  also   Huntington  in   Bulletin  of  the  American 
Mathematical  Society,  Vol.  IX,  p.  40.     H.  (irassmanu  {Lehrbuch  der  Arith- 
metik)  was  the  first  to  derive  7  and  8  from  tJ. 


THE   NATURAL   NUMBERS  13 

General  theorem  regarding  sums.     By  making  repeated  appli-    38 
cation  of  these  laws,  §§  34,  35,  it  can  be  shown  that 

The  sum  of  any  finite  number  of  numbers  will  be  the  same, 

whatever  the  order  in  which  we  arrange  them,  or  whatever  the 

manner  in  which  we  group  them,  when  adding  them. 

Thus,  a  +  b  +  c  +  d  =  a  +  c  +  b  +  d. 

For  a  +  b  +  c  +  d  =  a  +  (b  +  c)  +  d  §35 

=  a  +  {c  +  b)  +  d  §  34 

z=a  +  c  +  b  +  d.  §36 

Rules  of  equality  and  inequality  for  sums.     First.     From  the     39 
definition  of  sum,  §  29,  and  the  rules  of  §  24,  it  follows  that 

1.  If  a  =  b,  then  a  -{-  c  =  b  -]-  c. 

2.  If  a  <  b,  then  a  +  c  <  b  +  e. 

3.  If  a  >  b,  then  a  -{-  c  >  b  -^  c. 

Here  1  is  obvious,  since  it  a  =  b,  then  a  and  b  denote  the  same  number. 

We  may  prove  3  as  follows,  and  2  similarly. 

Ifa>5,  let  a  =  6  +  cZ,  §31. 

Then  a  +  c  =  (6  +  d)  +  c  =  {&  +  c)  +  d,  §§  34,  35,  .-,  >6  +  c. 

Second.     From  1,  2,  3  it  follows  conversely  that 

4.  li  a  -\-  c  =  b  -\-  c,  then  a  =  b. 

5.  If  a  +  c  <  b  +  c,  then  a  <  b. 

6.  U  a  -\-  c>  b  -\-  c,  then  a  >  b. 

Thus,  ii  a  +  c  =  b  +  c,  then  a  =  b. 

Foi-  otherwise  we  must  have  either  a<b  and  therefore  a  +  c<b  +  c 
;by  2),  or  else  a>b  and  therefore  a  +  Ob  +  c  (by  3). 

Third.     It  also  follows  from  1,  2,  3  that 

7.  If  a  =  b,  and  c  =  d,  then  a  -\-  c  =  b  -\-  d. 

8.  If  a  <  b,  and  c  <  d,  then  a  -]-  c  <  b  -{-  d. 

9.  If  a  >  b,  and  c  >  d,  then  a  -\-  c  >  b  -\-  d. 

Thus,  if  a  =  b,  then  a  +  c  =  6  +  c,  and  if  c  =  d,  then  b  +  c  =  b  +  d. 
Hence  a  +  c  =  b  -^  d. 


14  A    COLLEGE    ALGEBRA 


MULTIPLICATION 


40  Definition  of  multiplication.  To  multiply  a  by  5  is  to  find  the 
sum  of  b  numbers,  each  of  which  is  a. 

We  call  this  sum  the  product  oi  a  hj  b  and  express  it  in 
terms  of  a  and  ^  by  a  x  b,  oi  a- b,  or  simply  ab. 
Hence,  by  definition, 

41  ab  =  a  -\-  a  ■•  -to  b  terms. 

42  We  also  call  a  the  multiplicand,  b  the  multiplier,  and  a  and  b 
the  factoi's  of  ab. 

43  The  process.  To  find  products  by  repeated  addition  would 
be  very  laborious.  We  therefore  memorize  products  of  the 
smaller  numbers  (multiplication  tables),  and  from  these  derive 
products  of  the  larger  numbers  by  aid  of  the  laws  of  addi- 
tion and  the  laws  of  multiplication  explained  in  the  following 
sections. 

44  The  laws  of  multiplication.  Multiplication,  like  addition,  is 
a  commutative  and  an  associative  operation,  and  it  is  "dis- 
tributive "  with  respect  to  addition ;  that  is,  it  conforms  to 
the  following  three  laws  : 

45  The  commutative  law.     ab  =  ba. 

The  result  of  multiplying  a  %  b  is  the  same  as  that  of  mtd- 
tiplying  b  by  a. 

Thus,  2-3  =  0  and  3  •  2  =  G. 

46  The  associative  law.     a  (be)  =  (ah)  c, 

The  result  of  multiplying  a  by  the  jiroduct  be  is  the  same  as 
that  of  multiplying  the  product  ab  by  c. 

Thus,  2  (3  •  4)  =  2  •  12  =  24  ;  and  (2  •  3)4  =  6  •  4  =  24. 
In  practice  we  write  abc  instead  of  {ab)c.     Compare  §  36. 

47  The  distributive  law.     a(b  -\-  c)=  ab  +  ac, 

The    result   of  first   adding  h  and  c,    and    then   multiply- 
ing a  by  the  sum  so  obtained,   is  the  same  as  that  of  first 


THE    NATURAL   NUMBERS  15 

uiultiplymcj  a  hy  b  and  a  by  c,  and  then  adding  the  joroducts 
so  obtained. 

Thus,  3  (4  +  5)  =  3  •  9  =  27  ;  and  3  •  4  +  3  •  5  =  12  +  15  =  27. 

Proofs  of  these  laws.      We  may  prove  these  laws  as  follows :     48 

First.     T\\Q  distributive  law :  ab -{- ac  =  a(b  +  c).  (1) 

For  ah  +  ac  =  {a  +  a-\r  •  •  ■  toh  terms)  +  (a  +  «+■••  to  c  terms)   §  41 
=  a  +  a  +  a  +  •  •  •  to  (6  +  c)  terms  =  a  (6  +  c).      §§  35,  41 

Hence  a (1)  -\-  c  +  ■■■)=  ab  -\-  ac  -{-■■  -.  (2) 

Thus,  a{b  +  c  -\-  d)  =  a{h  ^-  c)  +  ad  =  ah  +  ac  +  ad.       by  (1)  and  § 35 

We  also  have  ac  +  he  =  (a.  +  b)  c.  (3) 

For  ac  +  hc  =  {a  +  a  +  ■  ■  ■  to  c  terms)  +  (h  +  b  +  ■  ■  ■  to  c  terms) 

=  (a  +  5)  +  (a  +  &')  +  •••  to  0  terms  =  {a  +  b)c.  §  38 

Second.      The  coniniutative  law:  ab  =  ba. 

a6  =  (1  -f  1  +  . . .  to  a  terms)  b 

=  1 .  6  +  1  •  5  H to  a  terms  by  (3) 

=  6  +  6  +  •  •  •  to  a  terms  =  ba.  §  41 

Third.      The  associative  law:   (ab)c  =^  a(bc). 

(ab)  c  =  ah  +  ab  +  ■■•  to  c  terms  §  41 

=  a{h  +  b  +  ■■■  toe  terms)  =  a  (be).       by  (2)  and  §  41 

General   theorem   regarding    products.     These    laws    can    be     49 
extended  to  products  of  any  finite  number  of  factors.     Thus, 

The  j)voduct  of  any  finite  number  of  factors  is  independent  of 
the  order  in  tvhich  the  factors  are  multiplied  tor/ether. 

Rules  of  equality  and  inequality  for  products.     These  are  :  50 

1.  If  a  =  b,  then  ac  =  be.  4.    If  ac  =  be,  then  a  =  b. 

2.  If  a  <  b,  then  ac  <  be.  5.    If  ac  <  be,  then  a  <b. 

3.  If  a  >  b.  then  ac  >  be.  6.    If  ac  >  be,  then  a  >  b. 


16  A   COLLEGE    ALGEBRA 

Here  1  is  obvious,  since  ii  a  =  b,  then  a  and  b  denote  the  same  number. 
We  may  prove  3  as  follows,  and  2  similarly. 

If  a>b,  let  a  =  b  +  d.     Then  ac  =  {b  +  d)c  =  bc  +  dc,  .:  >bc. 

The  rules  4,  5,  6  are  the  converses  of  1,  2,  3  and  follow  from  them  by 
the  reasoning  used  in  §  39. 

From  1,  2,  3,  by  the  reasoning  employed  in  §  39,  it  follows 
that 

If  a  —  b  and  c  =  d,  then  ac  =  hd. 

li  a  <h  and  c  <  d,  then  ac  <  bd. 

U  a>  b  and  c>  d,  then  ac  >  bd. 


II.     SUBTRACTION   AND    THE    NEGATIVE 

THE   COMPLETE   SCALE 

51  Subtraction.  To  subtract  3  from  5  is  to  find  what  number 
occupies  the  3d  place  before  5  in  the  natural  scale. 

We  find  this  number,  2,  by  counting  three  numbers  backward 
in  the  scale,  beginning  at  4,  thus  :  4,  3,  2. 

We  indicate  the  operation  by  the  sign,  — ,  read  "minus," 
writing  5  —  3  =  2. 

And,  in  general,  to  subtract  b  from  a  is  to  find  what  number 
occupies  the  ^th  place  before  a. 

We  call  this  number  the  remainder  obtained  by  subtracting 
b  from  a,  and  represent  it  in  terms  of  a  and  b  by  the  expres- 
sion a  —  b.     We  also  call  a  the  mimierid  and  b  the  subtrahend. 

52  Addition  and  subtraction  inverse  operations.  Olearly  the  third 
number  before  5  is  also  the  number  from  which  5  can  be 
obtained  by  adding  3. 

And,  in  general,  we  may  describe  the  remainder  a  —  b  either 
as  the  ^'th  number  before  a,  or  as  the  number  from  which  a 
can  be  obtained  by  adding  b,  that  is,  as  the  number  which  is 
defined  by  the  equation 

53  (^a-b)-\-b  =  a. 


SUBTRACTION    AND    THE   NEGATIVE  17 

Again,  since  saying  that  7  occupies  the  3d  place  after  4  is 
equivalent  to  saying  that  4  occupies  the  3d  place  before  7,  we 
have  4  +  3  —  3  =  4.     And  so,  in  general, 

{a -\- b)  -  b  =^  a.  54 

Since  a  +  ^  —  ^  =  a,  §  54,  subtraction  undoes  addition ;  and     55 
since  a  —  b  -\-  b  —  a,  §  53,  addition  undoes  subtraction.     We 
therefore  say  that  addition  and  subtraction  axQ  inverse  opera- 
tions. 

The  complete  scale.     The  natural  scale  does  not  fully  meet    56 
the  requirements  of  subtraction  ;  for  this  scale  has  Si  first  num- 
ber, 1,  and  we  cannot  count  backward  beyond  that  number. 

Thus,  on  the  natural  scale  it  is  impossible  to  subtract  4  from  2. 

But  there  are  important  advantages  in  being  able  to  count 
backward  as  freely  as  forward.  And  since  the  natural  scale 
is  itself  merely  a  system  of  signs  arranged  in  a  definite  order, 
there  is  no  reason  why  we  should  not  extend  it  backward  by 
placing  a  new  ordinal  system  of  signs  before  it. 

We  therefore  invent  successively  the  signs :  0,  which  we 
place  before  1 ;  —  1,  which  we  place  before  0 ;  —  2,  which 
we  place  before  —  1 ;  and  so  on. 

In  this  manner  we  create  the  complete  scale 

. . .,  _  5,  -  4,  -  3,  -  2,  -  1,  0,  1,  2,  3,  4,  5,  • .  •, 

which  has  neither  a  first  nor  a  last  sign  or  ''  number,"  and  on 
which  it  is  therefore  possible  to  count  backward,  as  well  as 
forward,  to  any  extent  Avhatsoever. 

Observe  the  sijmmetry  of  this  scale  with  respect  to  the  sign     57 
0.     As  3  is  the  third  sign  after  0,  so  —  3  is  the  third  sign 
before  0 ;  and  so  in  general. 

Meaning   of   the   new  numbers.     One    of   these    new   signs,     58 
namely  0,  may  be  said  to  have  a  cardinal  meaning.     Thus, 
counting  backward  from  3  corresponds  to  the  operation  of 
removing  the  elements  of  any  group  of  3  things,  one  at  a  time= 
This  operation  may  be  continued  until  all  the  elements  have 


18  A   COLLEGE    ALGEBRA 

been  removed,  and  we  may  call  0  the  sign  of  the  cardinal  num- 
ber of  the  resulting  "  group  "  of  no  elements.  We  therefore 
often  regard  0  as  one  of  the  natural  numbers. 

But  —  1,  —  2,  —  3,  •  •  •  have  no  cardinal  meaning  whatsoever. 

On  the  other  hand,  all  these  new  signs  have  the  same  ordinal 
character  as  the  natural  numbers.  Every  one  of  them  occupies 
a  definite  position  in  an  ordinal  system  which  includes  the 
natural  numbers  also.  And  we  may  consider  it  definedhy  this 
position  precisely  as  we  may  consider  each  natural  number 
defined  by  its  position  in  the  scale.  We  regard  this  as  a 
sufficient  reason  for  calling  the  signs  —  1,  —  2,  —  3  •  •  •  numbers. 

59  Positive  and  negative.     To  distinguish  the  new  numbers  —  1, 

—  2,  —  3,  •  •  •  as  a  class  from  the  old,  we  call  them  negative, 
the  old  jjositive. 

The  numbers  of  both  kinds,  and  0,  are  called  integers  to 
distinguish  them  from  other  numbers  to  be  considered  later. 

60  Algebraic  equality  and  inequality.  Let  a,  b,  c  denote  any  num- 
bers of  the  complete  scale.  According  as  a  precedes,  coincides 
with,  or  follows  b,  we  write  a  <  b,  a  =  b,  or  a  >  b. 

61  Since  by  definition  the  complete  scale  is  an  ordinal  system, 
§  17,  the  rules  of  §  24  apply  to  it  also ;  thus, 

If  a  <  b  and  b  <  c,  then  a  <  c. 

62  When  a  <  b,  that  is,  when  a  precedes  b  in  the  complete 
scale,  it  is  customary  to  say  that  a  is  algebraically  less  than  b, 
or  that  b  is  algebraicalli/  greater  than  a. 

Observe  that  the  words  "less"  and  "greater,"  as  thus  used,  mean 
"precede"  and  "follow,"  in  the  complete  scale  —  this  and  nothing  more. 
Thus,  "  —  20  is  less  than  —  18  "  means  merely  "  —  20  precedes  —  18." 

63  Absolute  or  numerical  values.     We  call  3  the  numerical  value  of 

—  3  or  its  absolute  value,  and  use  the  symbol  |  —  3 1  to  represent 
it,  writing  |  —  3 1  =  3.     Similarly  for  any  negative  number. 

The  numerical  value  of  a  positive  number,  or  0,  is  the  number 
itself.     Thus|3|  =  3. 


SUBTRACTION   AND   THE   NEGATIVE  19 

Numerical  equality  and  inequality.     Furthermore  we  say  of    64 
any  two  numbers  of  the  complete  scale,  as  a  and  b,  that  a  is 
numerically  less  than,  equal  to,  or  greater  than  b,  according 
as  |a|  <,  =,  or  >  |b|. 

Thus,  while  -  3  is  algebraically  less  than  2,  it  is  numerically  greater 
than  2,  and  while  -  7  is  algebraically  less  than  -  3,  it  is  numerically 
greater  than  —  3. 

OPERATIONS   WITH  NEGATIVE   NUMBERS 

New  operations.     We  also  invent  operations  by  which  the     65 
negative  numbers  and  0  may  be  combined  with  one  another 
and  with  the  natural  numbers,  as  the  latter  are  themselves 
combined  by  addition,  multiplication,  and  subtraction. 

We  call  these  operations  by  the  same  names,  and  indicate 
them  in  the  same  way,  as  the  operations  with  natural  numbers 
to  which  they  correspond. 

Employing  a,  as  in  §  60,  to  denote  any  number  of  the  com- 
plete scale,  but  a  and  b  to  denote  natural  numbers  only,  we 
may  define  these  new  operations  as  follows  : 

Definitions  of  addition  and  subtraction.     These  are :  66 

1.  a  +  &  is  to  mean  the  hth.  number  after  a. 

2.  a  —  i  is  to  mean  the  hi\\  number  before  a. 

3.  a  +  0  and  a  —  0  are  to  mean  the  same  number  as  a. 

4.  a  -f  (—  V)  is  to  mean  the  same  number  as  a  —  &. 

5.  a  —  (—  b)  is  to  mean  the  same  nmnber  as  a  +  6. 

In  other  words,  adding  a  positive  number  h  to  any  number 
a  is  to  mean,  as  heretofore,  counting  b  places  forward  in 
the  scale ;  subtracting  it,  counting  b  places  backward  :  while 
adding  and  subtracting  a  negative  number  are  to  be  equiva- 
lent respectively  to  subtracting  and  adding  the  corresponding 
positive  number. 


20  A   COLLEGE   ALGEBRA 

Thus,  byl,  -3  +  2  =  — 1,  since  —  1  is  the  2d  number  after  —  3. 

by  2,  2  —  5  =  —  3,  since  —  3  is  the  5th  number  before  2. 

by  4,  -  5  +  (-  2)  =  -  5  -  2  =  -  7  (by  2). 
by  5,  -  6  -  (-  2)  =  -  6  +  2  =  -  4  (by  1). 

67  Definition  of  multiplication.     This  is  : 

1.  0  •  a  and  a  ■  0  are  to  mean  0. 

2.  a  (—  b)  and  (—  a)  h  are  to  mean  —  ah. 
•3.    (—  a)  (—  b)  is  to  mean  ah. 

In  other  words,  a  product  of  two  factors,  neither  of  which  is 
0,  is  to  be  positive  or  negative  according  as  the  factors  have 
the  same  or  opposite  signs.  And  in  every  case  the  numerical 
value  of  the  product  is  to  be  the  product  of  the  numerical 
values  of  the  factors. 

Thus,  by  2,       3  x  -  2  =  -  6,  and  -  3  x  2  =  -  6. 
by  3,  -  3  X  -  2  =  6. 

68  The  origin  and  significance  of  these  definitions.  Observe  that 
the  statements  of  §§66  and  67  are  neither  assumptions  nor 
theorems  requiring  demonstration,  but  what  we  have  called 
them  —  definitions  of  neiv  operations. 

Thus,  it  would  be  absurd  to  attempt  to  prove  that  2  (—  3)  =  —  2  •  3  with 
nothing  to  start  from  except  the  definition  of  multiplication  of  natural 
numbers,  §  40,  for  the  obvious  reason  that  —  3  is  not  a  natural  number. 
The  phrase  "2  taken  —  3  times"  is  meaningless. 

But  why  should  such  operations  be  invented  ?  To  make 
the  negative  numbers  as  serviceable  as  possible  in  our  study 
of  relations  among  numbers  themselves  and  among  things  in 
the  world  about  us. 

The  new  operations  have  not  been  invented  arbitrarily  ;  on 
the  contrary,  they  are  the  natural  extensions  of  the  old  opera- 
tions to  the  new  numbers. 

In  dealing  with  the  natural  numbers,  we  first  defined  addi- 
tion as  2i  process  —  coimting  forward  —  and  then  showed  that 


SUBTRACTION    AND    THE   NEGATIVE  21 

the  results  of  this  process  have  two  properties  ivhich  are  inde- 
pendent of  the  values  of  the  numbers  added,  namely  : 

1.    a  +  b=^b  -\-  a.  2.    a  +  (Ij  +  c)  =  {a  +  b)  +  c. 

Similarly  we  proved  that  products  possess  the  three  general 
properties  : 

3.    ab  =  ba.     4.    a  (be)  =  (at)  c.     5.    a  (b  -\-  c)  =  ab  -{-  ac. 

When  we  employ  letters  to  denote  numbers,  these  properties 
1-5  become  to  all  intents  and  purposes  our  tvorkinr/  definitions 
of  addition  and  multiplication ;  for,  of  course,  we  cannot  then 
actually  carry  out  the  processes  of  counting  forward,  and  so  on. 

Clearly  if  corresponding  operations  with  the  new  numbers 
are  to  be  serviceable,  these  "  definitions "  1-5  must  apply 
to  them  also.  And  §§  66,  67  merely  state  the  solution  of 
the  problem  : 

To  make  such  an  extension  of  the  meanings  of  addition,  multi- 
plication, and  subtraction  that  sums  and  products  of  any  num- 
bers of  the  comjilete  scale  may  have  the  properties  1-5,  and  that 
subtraction  may  contimie  to  be  the  inverse  of  addltioii. 

Thus,  (1)  when  we  define  adding  a  positive  number  &  to  a  as  counting 
forward,  and  subtracting  it  as  counting  backward,  we  are  merely  repeating 
the  old  definitions  of  addition  and  subtraction. 

(2)  From  this  definition  of  addition  it  follows  that  —6  +  6  =  0. 

But  if  the  commutative  law  a  +  6  =  6  +  aisto  hold  good,  we  must  have 
—  6  +  6  =  6  +  (—  6),  and  therefore  6  +  (—  6)  =  0 ;  or,  since  6  —  6  =  0, 
we  must  have  b  +  ( —  6)  =  6  —  6. 

This  suggests  the  definition  a  +  ( —  6)  =  a  —  6. 

(3)  If  our  new  addition  and  subtraction  are,  like  the  old,  to  be  inverse 
Operations,  we  must  also  have,  as  in  §  66,  5,  a  —  (—  6)  =  a  +  6. 

(4)  Again,  to  retain  the  old  connection  between  addition  and  multipli- 
cation, §  41,  we  must  have,  as  in  §  67,  2, 

( —  a)  6  =  —  a  +  ( —  a)  +  •  •  ■  to  6  terms 

=  —  a  —  a  —  ■  ■  ■  to^  lisrms  =  —  ah. 

(6)  If  the  commutative  law  ah  =  ba  is  to  hold  good,  we  must  also  have 
a(- 6)  =  (- 6)a  =  -6a  =  -a6,  as  in  §67,  2, 


22  A   COLLEGE    ALGEBRA 

(6)  Similarly,  0  +  0  +  ■  •  •  to  a  terms  =  0,  and  this  fact  together  with 
our  wish  to  conform  to  the  law  ab  =  ba  leads  to  the  definitions  of  §  67,  1, 
namely,  0  •  a  =  0  and  a  •  0  =  0. 

(7)  Finally,  it  follows  from  (6)  that  {- a)  {- b  +  b)  =  - a -0  =  0. 
But   if   the   distributive   law   is  to   hold   good,    we   must   also    have 

{-  a){-  b  +  b)  =  (-  a)  {  ~  b)  +  {-  a)b  =  (-  a)  (-  b)  -  ab,  by  (4). 

"We  therefore  have  {—  a)  {—  b)  -  ab  =  0.  And  since  also  ab  —  ab  =  0, 
we  are  thus  led  to  define  (—  a)  (—  b)  as  ab,  as  in  §  67,  3. 

69  The  operations  just  defined  conform  to  the  commutative,  asso- 
ciative, and  distributive  laws.  It  remains  to  prove  that  the  new- 
operations  are  in  complete  agreement  with  the  laws  which 
suggested  them. 

To  begin  with,  we  have 

a  +  (6  +  o)  =  a  +  &  +  c,  (1) 

a-(b  +  c)  =  a-b-c,  (2) 

a  +  b-b=SL-b  +  b  =  ai,  (3) 

as  follows  from  the  definitions  of  addition  and  subtraction  as 
counting  forward  and  backward,  by  the  reasoning  in  §§  37,  52. 

I.  The  commutative  law,  a  +  b  =  b  +  a. 

First,  -a  +  b  =  b+(-a). 

Forifa>6,  let  a  =  d  +  b.  §§31,34 

Then  -a  +  b  =  -{d  +  b)  +  b 

=,^d-b  +  b  =  -d;  by  (2)  and  (3) 

and  b  +  {-a)  =  b-{b  +  d),  §  66,  4 

=  b-b-d  =  -d.  by  (2) 

Proceed  in  a  similar  manner  when  b>a. 
Second,  -  a  +  (- b)  =  - b -{- (-  a). 
For -o  if  {-?>)  = -(a +  &)  =  -(&  +  «)  = -6 +  (-«),  by  (2)  and  §66,4. 

II.  The  associative  law,  a  +  (b  +  c)  =  (a  +  b)  +  c. 
First,  8i+[b+(-c)^  =  a  +  b+(-  c). 

Forif6>c,  let  b  =  d  +  c.  §§31,34 


SUBTRACTION   AND   THE   NEGATIVE  23 

Then     a  +  [6  +  (-c)]  =  a +[d  +  c +(- c)]  =  a  +  d, 
and  a +  fe  + (- c)  =  a  +  d  +  c  +  (- c)    =  a  +  d.  by  (3) and  §66,4 

Proceed  in  a  similar  manner  when  c>b. 

Second,  a  +  [(-  &)  +  c]  =  a  +  (-  6)  +  c. 

This  follows  from  I  and  the  case  just  considered. 

Third,  a  +  [-  Z-  +  (-  c)]  =  a  +  (-  i)  +  (-  c). 

This  follows  from  (2)  and  §  66,  4,  since  -  6  +  (-  c)  =  -  (6  +  c). 

III.  The  commutative  law,  ab  =  ba. 
First,  (-  a)b^b(-  a). 

For  {-a)b  =  -ab  =  -~ba==b{-a).  §45;  §67,2 

Second,  (-  a)  (-  b)  =  (-  b)  (-  a). 

For  {-a){-b)  =  ab  =  ba^{~b){-a).  §45;  §67,3 

IV.  The  associative  law,  a  (be)  =  (ab)c. 

First,  (-  a)  [(-  b)  (-  c)]  =  [(-  a)  (-  ^)]  (-  c). 

For  {-a)[{-b){~c)]  =  {-a)-bc  =  -abc,    §  46  ;  §  67,  2,  3 

and  [(_a)(_6)](_c)  =  aft.  (- c)  =-a&c.  §67,2,3 

Second,  the  other  cases  may  be  proved  in  the  same  way. 

V.  The  distributive  law,  a  (b  +  c)  =  ab  +  ac. 
First,  a[b  +(—  c)]  =  ab  +  a  (—  c). 

For  [b  +{-  c)]a  =  [b  +  {-  c)]  +  [b  +  {-  c)]  +  ■■■  to  a  terms 

=  6  +  6-1 to  a  terms  +  ( —  c)  +  ( -  c)  +  •  •  •  to  a  terms 

=  ba  +  {-c)a.  §  41 ;  §  67,  2  ;  II  and  III 

Hence  a[b  +  {- c)]  =ab  +  a(- c)  by  III 

Second,  from  this  case  the  others  readily  follow. 

Thus,       (-a)[6+(-c)]=-a[6  +  (-c)] 

=  -Qab  +  a(-  c)]  =  (-  a)b+{-  a)  (-  c). 


24  A   COLLEGE   ALGEBRA 

70  The  general  result.  As  has  already  been  observed,  §  68,  in 
literal  arithmetic  or  algebra,  the  laws  a  +  b  =  b  +  a,  and  so 
on,  are  equivalent  to  dejinitions  of  addition  and  multiplication, 
even  when  the  letters  a,  b,  c  denote  natural  numbers.  And  we 
have  now  shown  that  these  definitions  apply  to  all  numbers 
of  the  complete  scale. 

By  means  of  these  laws  we  may  change  the  form  of  a  literal 
expression  without  affecting  its  value,  whatever  numbers  of 
the  complete  scale  the  letters  involved  in  the  expression  may 
denote. 

Thivs,  whether  a,  6,  c,  d  denote  positive  or  negative  integers,  we  have 
(a  +  6)  (c  +  d)  =  (a  +  6)  c  +  (a  +  f")  d 
=  ac  +  hc  +  ad  -\-  bd. 

71  Rules  of  equality  and  inequality  for  sums.  We  may  prove  by 
reasoning  similar  to  that  in  §  39  that 

According  as  a  < ,  = ,  or  >  b, 

so  is  a  +  c  < ,  = ,  or  >  b  +  c ; 

and  conversely. 

Hence  it  is  true  for  positive  and  negative  numbers  alike  that 

72  An  equation  remains  an  equation  and  the  sense  of  an  inequality 
remains  unchanged  when  the  same  number  is  added  to  both  sides, 
or  is  subtracted  from  both. 

73  Rules  of  equality  and  inequality  for  products.  Observe  that 
changing  the  signs  of  any  two  numbers  a  and  b  reverses  the 
order  in  which  they  occur  in  the  complete  scale,  §  57. 

Thus,  we  have  -  3  <  -  2,  but  3  >  2  ;   -  5  <  2,  but  5  >  -  2. 

From  this  fact  and  the  reasoning  of  §  50  it  follows  that 
According  as  a  <,  = ,  or  >  b, 

so  is  ac  < ,  = ,  or  >  be, 

but  a(— c)  >,  =,  or  <  b(— c); 

and  conversely.     Hence 


SUBTRACTION    AND    THE    NEGATIVE  25 

Mxdtiphjing  both  sides  of  an  equation  by  the  same  number,     74 
positive  or  negative,  leaves  it  an  equation. 

Multiplying  both  sides  of  an  inequality  by  the  same  positive 
number  leaves  its  sense  unchanged. 

But  multiplying  both  sides  of  an  inequality  by  the  same  nega- 
tive number  changes  its  sense,  from  <.to>,  or  vice  versa. 

From  the  first  of  these  rules  and  the  definition  of  multipli- 
cation by  0,  namely  a  •  0  =  0,  we  derive  the  following  important 
theorem  : 

1.  If   a  =  b,    then  ac  =  be.  75 

2.  If  ac  =  be,  then    a  =  b,  uitless  c  =  0. 

The  exceptional  case  under  2  should  be  carefully  observed. 
Thus,  from  the  true  equation,  2  •  0  =  3  •  0,  of  course  it  does  not  follow 
that  2  =  3. 

Zero  produets.     If  a  j^roduct  be  0,  one  of  its  factors  must  be  0.     76 

Thus,  if  ab  =  0,  either  a  =  0  or  b  =  0. 

For,  since  0  ■  b  is  also  equal  to  0, 
we  have  ab  =  0  •  b, 

and  therefore  a  =  0,  unless  b  =  0.  §  75 

Numerieal  values  of  products.     The  numerical  value  of  a  prod-     77 
uct  of  two  or  more  factors  is  the  product  of  the  numerical 
values  of  the  factors. 

Thus,  |(-2)(-  3)(-4)|  =  |~24|=24;  and|-2|.|-3|.|-4|  =  24. 

Numerical  values  of  sums.     The  numerical  value  of  a  sum  of     78 
two  numbers  is  the  stmt  of  their  numerical  values  when  the 
numbers  are  of  like  sign,  but  the  numerical  difference  of  these 
values  when  the  numbers  are  of  contrary  sign. 

Thus,  |-3  +  (-5)|  =  |-8|  =  8;and|-3|-M-5|  =  3-)-5  =  8. 
But         |2-l-(-5)|  =  |-3|  =  3;  and|-5|-2  =  3. 


26  A   COLLEGE   ALGEBRA 


THE  USE  OF   INTEGRAL   NUMBERS   IN  MEASUREMENT 

79  Measurement.  We  use  numbers  not  only  to  record  the 
results  of  counting  groups  of  distinct  things,  but  also  to  indi- 
cate the  results  of  vieasur'ing  magnitudes,  such  as  portions  of 
time,  straight  lines,  surfaces,  and  so  on. 

80  We  measure  a  magnitude  by  comparing  it  with  some  particu- 
lar magnitude  of  the  same  kind,  chosen  as  a  unit  of  measure. 

81  If  the  magnitude  contains  the  unit  a  certain  number  of 
times  exactly,  we  call  this  number  its  measure. 

In  particular,  we  call  the  measure  of  a  line  segment  the 
length  of  the  segment. 

Thus,  we  may  measure  a  line  segment  by  finding  how  many  times  we 
can  lay  some  chosen  unit  segment,  say  a  foot  rule,  along  it. 

If  we  find  that  it  contains  the  foot  rule  exactly  three  times,  we  say  that 
it  is  three  feet  long,  or  that  its  length  —  that  is,  its  measure  —  is  3. 

82  The  usefulness  of  the  natural  numbers  in  measurement  is 
due  to  the  fact  that,  by  their  relative  positions  in  the  natural 
scale,  they  indicate  the  relative  sizes  of  the  magnitudes  whose 
measures  they  are. 

83  Application  of  the  negative  numbers  to  measurement.  We  often 
have  occasion  to  make  measurements  in  opposite  "  directions  " 
from  some  fixed  "  point  of  reference." 

Thus,  we  measure  time  in  years  before  and  after  the  birth  of  Christ, 
longitude  in  degrees  west  and  east  of  Greenwich  or  Washington,  tempera- 
ture in  degrees  helow  and  above  zero. 

We  may  then  distinguish  measurements  made  in  the  one 
direction  from  those  made  in  the  other  by  the  simple  device 
of  representing  the  one  by  positive  numbers,  the  other  by 
negative  numbers. 

84  Thus,  consider  the  following  figure  : 


-P_4       P-s       P-i       /•_!        O  Pi  P.^  Ps 


DIVISION    AND    FRACTIONS  27 

Here  the  fixed  point  of  reference,  or  origin,  is  0,  the  unit 
is  OPi,  and  the  points  P^,  -P3,  •••,  P-i,  -P_2j  •••  are  such  that 
OPi  =  P1P2  =  P2P3  =  ■••  =  P^iO  =  P_2P-i  =  •■■• 

Above  these  points  we  have  written  in  their  proper  order  the 
numbers  of  the  complete  scale,  so  that  0  comes  over  0. 

The  distance  of  each  point  P  from  0,  —  that  is,  the  length 
of  the  segment  OP,  —  is  then  indicated  by  the  numerical  value 
of  the  number  written  above  it ;  and  the  direction  of  P  from  O 
is  indicated  by  the  sign  of  that  number. 

Thus,  —  3  over  P_  3  indicates  that  P-3  is  distant  3  units  to  the  left  of  O. 

Moreover,  the  order  in  which  the  points  occur  on  the  line  is 
indicated  by  the  order  in  which  the  corresponding  numbers 
occur  in  the  scale. 

Points  used  to  picture  numbers.     Inasmuch  as  there  is  a  one-     85 
to-one  relation,  §  2,  between  the  system  of  points  •  •  •,  P_2,  P_i, 

0,  Pi,   Po,  •  •  •    and  the  system  of   numbers  ■••,  — 2,  —1,   0, 

1,  2,  •••,  either  system  may  be  used  to  represent  the  other. 
In  what  follows  we  shall  frequently  use  the  points  to  picture 
the  numbers. 

III.     DIVISION   AND   FRACTIONS 

DIVISION  REPEATED   SUBTRACTION 

The  two  kinds  of  division.     There  are  tivo  operations  to  which     86 
the  name  division  is  applied  in  arithmetic  and  algebra.     The 
one  may  be  described  as  repeated  subtraction,  the  other  as  the 
inverse  of  multiplication.     There  is  a  case  in  which  the  two 
coincide.     We  call  this  the  case  of  exact  divisio7i. 

Division  repeated  subtraction.     To  divide  7  by  3  in  the  first     87 
of  these  senses  is  to  answer  the  two  questions  : 

1.  What  multiple  of  3  must  we  subtract  from  7  to  obtain  a 
remainder  which  is  less  than  3  ? 

2.  What  is  this  remainder  ? 


28  A   COLLEGE   ALGEBRA 

We  may  find,  the  answer  to  both  questions  by  repeatedly 
subtracting  3.  Thus,  since  7  —  3  =  4  and  4  —  3  =  1,  we  must 
subtract  3  twice,  or,  what  comes  to  the  same  thing,  we  must 
subtract  ^  X  2.     And  the  remainder  is  1. 

This  kind  of  division,  then,  is  equivalent  to  repeated  suhtrao 
tion.  Its  relation  to  subtraction  is  like  that  of  multiplication 
to  addition. 

Observe  that  the  four  numbers  7,  3,  2,  1  are  connected  by 
the  equation  7  =  3-2  +  1 

And  so  in  general,  if  a  and  b  are  any  two  natural  numbers,  to 
divide  a  by  b,  in  the  sense  now  under  consideration,  is  to  find 
two  natural  numbers,  q  and  r,  one  of  which  may  be  0,  such  that 

88  a  =  bq  -{-  7'  and  r  <  b. 

89  We  call  a  the  dividend,  b  the  divisor,  q  the  quotient,  and  r 
the  remainder. 

90  Note.  When  a  and  h  are  given,  two  numbers  q  and  r  satisfying  §  88 
may  always  be  found. 

Thus,  if  a  <  6,  we  have  q'  =  0  and  r  =  a. 

If  a>6,  it  follows  from  §§  31,  35  that  we  can  continue  the  sum 
6  +  5  +  •  ■  •  until  it  either  equals  a  or  will  become  greater  than  a  if  we  add 
another  b.  And  if  q  denote  the  number  of  terms  in  this  sum,  we  shall 
have,  §  41,  either  a  =  hq,  ov  a  =  bq  -{-  r,  where  r<b. 

Again,  when  a  and  b  are  given.,  but  one  pair  of  numbers  q  and  r 
satisfying  §  88  exists. 

For  were  there  a  second  such  pair,  say  q',  r',  we  should  have 

bq  +  r  =  bq'  +  r',  and  therefore  b{q  —  q')  =  r'  —  r. 

But  this  is  impossible,  since  r'  —  r  would  be  numerically  less  than  6, 
but  b{q  —  q')  not  numerically  less  than  b. 

91  Exact  division.  If  the  dividend  a  is  a  multiple  of  the  divisor 
b,  as  when  a  =  12  and  b  =  3,  the  remainder  r  is  0.  We  then 
say  that  a  is  exactly  divisible  by  b.  In  this  case  the  equation 
of  §  88  reduces  to  a  =  bq,  or 

92  qb  =  a. 


DIVISION    AND    FRACTIONS  29 

Hence  when  a  is  exactly  divisible  by  b,  we  may  also  define  the  93 
quotient  q  as  the  7iumber  which,  multiplied  bt/h,  ivill  jjroduce  a. 

In  this  case,  furthermore,  we  may  indicate  the  division  94 
thus,  a  -i-  b,  and  express  the  quotient  q  in  terms  of  a  and  b  by 

one  of  the  symbols  j  or  a/b,  writing  q  =  j  as  well  as  qb  =  a. 

THEOREMS   AND  FORMULAS   RESPECTING  EXACT   DIVISION 

Theorem  1.  Exact  divisioii  and  multiplication  are  inverse  95 
operations  ;  that  is, 

a  ^  b  X  b  =  a,  a7id  a  x  b  h-  b  =  a. 
These  equations  follow  from  the  definitions  hi  §  93  and  §  87  respectively. 

Theorem  2.  When  division  is  eocact,m}(Uiplying  dividend  and  96 
divisor  by  the  same  number  leaves  the  quotie7it  unchanged. 

For  if  a  —  qb,  then  am  =  q  ■  bin.  §§  50,  4G 

That  is,  if  g  =  ^,   then       g  =  — .  §94 

b  bin 

Theorem  3.  Exact  division,  like  ^nultijilication,  is  distributive  97 
ivith  respect  to  addition  and  sid>traction  ;  that  is, 

a      b      a  +  b  .a      b      a  —  b 

— I —  = }  and •  -  = 

c       c  c  c       c  c 

For  if  a  =  qc,  and  b  =  q'c, 

we  have  a  +  b  =  qc  +  q'c  =  {q  +  q')  c.  §§  39,  47 

a  -\-  b  ,      a      b  „  ^, 

Hence  =  g  +  (^'  =  -  +    .  §  94 

c  c      c 

And  similarly  for  subtraction. 

Thus,  ^  +  ^  =  6  +  3  =  9;   and  1^  +  ^  =  2-1  =  9. 

3       3  3  3 

Formulas  for  adding  and  subtracting  quotients.     These  are  98 

a       c       ad  -\-  be      a       c       ad  —  be 
■      b^d^       bd       '    b~d^  ~~bd 


30  A   COLLEGE   ALGEBRA 

And  similarly  for  subtraction. 

18   ,10      a  ,   o      Q  .  18- 5 +  10 -3      120      ^ 

Thus, =  6  +  2  =  8;  and  = =  8. 

3        5  8-5  15 

99         Formula  for  multiplying  quotients.     This  is 

a    c       ac 

h'd^bd' 

For  if  a  =  §&,  and  c  =  g'd,  we  have  ac  =  qq'  ■  bd.  §§  50,  45,  46 

^  o.   c  ,      ac  „  ^ . 

Hence  -.-  =  g.g=--.  §94 

0   a  bd 

15   6      ^    ^      ,^         ^.  15-6      90      ,^ 

Thus,  =  5-3  =  15;  and  =  —  =  15. 

3     2  '3-2        6 

100         Formula  for  dividing  one  quotient  by  another,  when  this  division 
is  exact.     This  is 


a       c       ad      ad 
h   '  d      he       be 


For  if  a  =  5&,  c  =  q'd,  and  also  q  =  q"q\ 

we  have  t  ^  -  =  <! -^  <l'  =  (l'\  §  94 

0      a 

and  ^  =  i^  =  ^  =  9".  §§96,94 

be       bq  a      q 

r^  24      10      ,       „      „  ,  24-5      120      „ 

Thus,  — -; =  4h-2  =  2;  and  =  —  =2, 

'  6        5  6  •  10       00 

101  Exact  division  for  negative  numbers.  Evidently  the  definition 
of  quotient  given  in  §  93  has  a  meaning  for  negative  numbers 
also,  whenever  the  numerical  value  of  the  dividend  is  exactly 
divisible  by  that  of  the  divisor.  Expressing  these  quotients 
as  in  §  94,  we  have  the  following  theorem  : 

102  Theorem  4.     If  a,  is  exacthj  divisible  by  b,  then 

—  a  a  a  a  — a a 

"b~"'~b'  ^"""b'  ^"b" 


DIVISION    AND   FRACTIONS  31 

For  if  a  =  g6,  we  have        —a  =  {—q)b.  §§73,67 

—  o,  (X  „  ^ . 

Hence  —  —  q  = §94 

b  h 

And  similarly  in  the  other  cases. 

Zero  in  relation  to  exact  division.     1.    On  the  other  hand,  the     103 
definition  of  quotient  given  in  §  93  is  meaningless  when  the 
divisor  is  0. 

For  q  X  0=  0,  no  matter  what  number  q  may  denote.  Hence  (1)  every 
number  is  one  which  multiplied  by  0  gives  0  ;  and  (2)  there  is  no  number 
which  multiplied  by  0  will  give  a. 

In  other  words,  according  to  the  definition  of  §  93  and  §  94,  the  Symbol 
0/0  would  denote  every  number,  and  a/0  no  number  whatsoever. 

2.  But  when  the  dividend  is.O,  and  the  divisor  some  num- 
ber h  which  is  not  0,  the  definition  of  §  93  has  a  meaning.  In 
fact,  the  quotient  denoted  by  Q /h  is  0. 

For,  according  to  §  94,  0/6  should  denote  the  number  which  multiplied 
by  b  gives  0  ;  and  0  is  that  number  (and  the  only  one),  since  0-6  =  0. 

FRACTIONS.     DIVISION   THE   INVERSE   OF   MULTIPLICATION 

The  second  kind  of  division  mentioned  in  §  86  is  the  gener- 
alization of  exact  division  defined  as  in  §  93.  It  requires  that 
fractions  be  introduced  into  the  number  system.  We  seek  an 
ordinal  definition  of  these  new  numbers,  like  that  given  the 
negative  numbers  in  §  56.  One  is  suggested  by  the  following 
theorem,  in  which  a,  b,  c,  d  denote  natural  numbers. 

Theorem  5.      Whe7i  a  is  exactly  divisible  byh,  and  c  by  d,  the     104 
quotients  a/b  and  c/d  occur  i^i  the  natural  scale  in  the  same 
relative  order  as  the  products  ad  and  be  /  that  is, 

a/b  <,  =,  or  >  c/d,  according  as  ad  <,  =,  or  >  be. 

1.    For  if  -  =  -,  then  -6d  =  -d6.  §50 

0      d  b  d 

But  -6  =  a,  and  -d  =  c.  §§93,94 

Hence  ad  =  6c. 


32  A   COLLEGE   ALGEBRA 

And  we  can  show  in  a  similar  manner  tliat 

K  a/h<c/d,  then  ad<hc;  and  if  a/h>c/d,  tlien  ad>bc. 

2.    But  from  all  this  it  follows,  conversely,  that 

If  ad  =  6c,  then  a/h  =  c/d. 

For  otherwise  we  should  have  either  (1)  a/h<c/d,  and   therefore 
ad<he,  or  (2)  a/h>c/d,  and  therefore  ad>bc. 
And  we  can  show  in  the  same  way  that 

If  ad<hc,  then  a/h<c/d;  and  if  ad>bc,  then  a/b>c/d. 

105  Enlarging  the  ordinal  number  system.  But  the  relative  ordei 
of  ad  and  be  in  the  scale  is  known,  whether  the  values  assigned 
a,  b,  c,  d  be  such  as  make  a  divisible  by  b,  and  c  by  d,  or  not. 

Therefore,  take  any  two  natural  numbers,  a  and  b,  of  which 

b  is  not  0,  and  with  them  form  the  expression   j,  or  a/b. 

If  a  is  exactly  divisible  by  b,  let  a/b  denote,  as  heretoforCj 
the  natural  number  which  is  the  quotient  of  a  by  i ;  but  if 
not,  regard  a/b  for  the  moment  merely  as  a  new  symbol,  read 
"  a  over  &,"  whose  relation  to  division  is  yet  to  be  given,  §  122. 

Then  give  to  all  such  symbols  a/b,  c/d,  and  so  on,  the 
property  of  order  already  possessed  by  those  which  denote 
natural  numbers,  by  supposing  them  arranged  in  accordance 
with  the  rule:  a/b  shall  precede,  coincide  with,  or  follow  c /d^ 
according  as  ad  precedes,  coincides  with,  or  follows  be. 

Or,  employing  the  signs  <,  =,  >,  as  heretofore,  to  mean 
"  precede,"  ''  coincide  with,"  "  follow,"  — 

106  Let  a/b  <,  =,  or  >  c/d,  according  «s  ad  <,  =,  or  >  be. 

Thus,  4/5  is  to  precede  7/8,  that  is,  4/5<7/8,  since  4-8<7-5. 
Again,  2/3  is  to  lie  between  0  and  1,  or  0<2/3<l.  For  0/l<2/3, 
since  0-3<2-l;  and  2/3  <  1/1,  since  2  •  1<  1  •  3. 

107  To  such  of  the  symbols  a/b  as  denote  natural  numbers 
this  rule  assigns  their  proper  places  in  the  scale  itself ;  while 
to  the  rest  it  assigns  places  between  consecutive  numbers  of 
the  scale. 


DIVISION   AND    FRACTIONS  33 

Note.     To  find  the  place  thus  given  any  particular  symbol  a/b  with     108 
respect  to  the  numbers  of  the  scale,  we  have  only  to  reduce  a  to  the  form 
a  =  bq  +  r,  where  r  <  6,  §  88.     Then  if  r  =  0,  so  that  a  —  bq,  our  rule 
makes  a/b  coincide  with  q.     But  if  r  is  not  0,  our  rule  places  a/b  between 
q  and  q  +  I. 

The  entire  assemblage  of  symbols   a/b  thus  defined   and     109 
arranged  —  like  the  natural  scale  which  forms  part  of  it  — 
is  an  ordinal  system. 

For  it  has  all  the  properties  of  an  ordinal  system  which  were  enumerated 
in  §§  17,  18. 

Thus,  if  a/b<c/d,  and  c/d<e/f,  then  a/b<e/f. 

For  if  a/b<c/d,  and  c/d<e/f, 

we  have  ad<bc,         and  cf<ed.  §106 

Multiplying  the  sides  of  the  first  of  these  inequalities  by  the  corre- 
sponding sides  of  the  second,  we  have 

§50 
§60 

§  106 

Fractions.     When  a/b  does  not  denote  a  natural  number,     110 
we  call  it  a,  fraction ;  and  we  call  a  its  numerator,  b  its  denomi- 
nator, and  both  a  and  b  its  terms.     Hence 

A  fraction  is  a  symbol  of  the  form  a/b,  defined  by  its  j^osition 
in  an  ordinal  system  which  includes  the  natural  numbers. 

From  an  ordinal  point  of  view,  therefore,  we  are  justified  in 
calling  fractions  numbers.* 

*  The  rule  of  §  106  may  also  be  used  to  define  symbols  of  the  form  1  /O,  2/0, 
and  so  on,  ordinally. 

Thus,  by  the  rule,  1  /O  will  follow  every  number  a/b  whose  denominator  h 
is  not  0.     For  l/0>a/6.  since  1  ■  6>a-0. 

Again,  1/0,  2/0.  and  so  on,  will  occupy  the  same  place  in  our  ordinal  system. 
For  1/0=2/0,  since  10=  2   0. 

But  the  rule  will  give  no  definite  position  to  the  symbol  0/0.  For  whatever 
the  values  of  a  and  b,  we  should  have  0/0=  a/b,  since  0  •  6  =  a  •  0. 


adcf<bced. 

Hence 

af<be, 

and  therefore 

a/b<e/f. 

34  A   COLLEGE   ALGEBRA 

111  Negative  fractions.  We  also  form  fractions  whose  numer- 
ators,   denominators,    or    both,     are     negath-e    integers,     as 

——5  — 7'  — T'  d^fi^iiig  them  ordinally  as  follows: 

2.  Every  negative  fraction  shall  x>recede  Q. 

3.  Negative  fractions  shall  be  arranged  tvith  respect  to  one 
another  {and  negative  integers)  in  accordance  with  the  rule: 

—  -<,=,  or  >  —  T'  according  as  —  ad  <,  =,  or  >—  be. 

112  The  system  of  rational  numbers.  To  distinguish  integers  and 
fractions  alike  from  other  numbers  which  we  have  yet  to  con- 
sider, we  call  them  rational  numbers.  And  we  call  the  system 
which  consists  of  all  these  numbers  the  rational  system. 

This  system  possesses  an  important  property  which  does 
not  belong  to  its  part,  the  integral  system,  namely: 

113  The  rational  system  is  dense;  t/iat  is,  betiveen  every  two 
unequal  rational  numbers  there  are  other  rational  numbers. 

For  let  -  and  -  be  any  two  fractions,  such  that     <  -  •     We  can  prove 
h  d  0     d 

as  follows  that  the  fraction  — lies  between  -  and  -  • 

2bd  b  d 

Since  -  <-,  we  have  ad<bc.  §  106 

b     d 

1.  If  we  add  ad  to  both  sides  of  ad<bc,  we  have,  by  §§  39,  50,  106, 

a    be  +  ad 

2ad<bc  +  ad,  .:  a{2bd)<b{bc  +  qd),  ••■-<    ^^^^    • 

2.  If  we  add  be  to  both  sides  of  ad<bc,  we  have  similarly 

bc  +  ad<2bc,  .-.  {be  +  ad)d<c(2bd),  ""YbdT^d' 

^.       V.  •'i       ^  5         ,         4 .  5  +  3  •  (>      38      10 

Thus,  between  -  and  -  we  have  —  —  TE,~  W^i' 

4  6  5s  •  4  •  o  4o       Zi 


DIVISION    AND   FRACTIONS  35 

Hence,  when  speaking  of  rationals,  one  must  carefully  avoid     114 
such  expressions  as  the  "  next  number  greater  or  less  "  than  a 
given  number ;   for  no  such  number  exists.     To  each  integer 
there  is  such  a  next  integer,  but  between  any  rational  and  a 
rational  assigned  as  the  next,  there  are  always  other  rationals. 

Operations  with  fractions.     In  what  follows  let  a,  b,  c,  d  denote     115 
any  given  integers,  positive  or  negative. 

In  §§  98-102  we  proved  that,  when  a/h  and  c/d  denote 
mtegers,  we  have 


^     a   ^    c       ad  +  he 
'    b^  d~       bd 

2    -- 
b 

e       ad  —  be 
d           bd 

a    c          ae 
h   d         he' 

-r 

e       ad       . 
f-  -  =  — ,  whe 
d       be 

ad  . 

—  is  an  integer. 

But  the  second  member  of  each  of  the  equations  1,  2,  3,  4 
has  a  meaning  even  when  a/b  and  c/d  are  not  integers.  Each 
of  them  is  a  definite  fraction  of  the  kind  defined  in  §§  110,  111. 

Hence  1,  2,  3,  4  at  once  suggest  an  extension  of  the  mean- 
ings of  addition,  subtraction,  multiplication,  and  division  which 
will  make  these  operations  applicable  to  fractions,  namely : 

The  sum  of  two  fractions  a/b  and  c/c?  is  to  mean  the  fraction     116 
{ad  +  he)/hd. 

The  difference  obtained  by  subtracting  the  fraction  e /d  from     117 
the  fraction  a/b  is  to  mean  the  fraction  (ad  —  bc)/hd. 

The  product  of  two  fractions  a/b  and  c/d  is  to  mean  the     118 
fraction  ac /bd. 

The  quotient  resulting  from  dividing  the  fraction  a/b  by  the     119 
fraction  c/d  is  to  mean  the  fraction  ad /he 

Observe  that  these  definitions  are  equivalent  to  tlie  rules  for  reckoning 
with  fractions  given  in  elementary  arithmetic. 

The  commutative,   associative,   and  distributive   laws   control     120 
these  generalized  operations. 

Thus,  a    c^oc^ca^c    a 

b   d     hd     dh     d   h  ^^       ' 


36  A   COLLEGE    ALGEBRA 

121  The  rules  of  equality  and  inequality,  §§  71,  73,  also  hold  good 
for  these  operations. 

Thus,  if  =  -;  •  ^'  then         v  =  :^  • 

b  f     d  f  b      d 

For  if  ^.f  =  :^.^  then  aedf=cebf.      §§118,100,111 

b  f     d  f 

Hence  ad  =  cb,         and  therefore -  =  - •  §§73,106,111 

0      a 

122  Definition  of  a  fraction  as  a  quotient.  The  fraction  a/b  may 
now  be  described  as  the  number  which  multijjlied  by  b  tc ill  pro- 
duce a,  that  is,  as  the  number  which  is  defined  by  the  equation 

123 


b 

■b  =  a. 

a 
b 

b-- 

a 

~b 

b  _ 
1  " 

ab 
"  \-b 

a 

T 

b 
b 

For  'l.b^l--  =  ^  =  ^-  =  a.  §§106,111,118 

b  6    1       l-b      \    b 

124  Division  the  inverse  of  multiplication.     From  §§  118,  119,  it 

follows  that 

area        ^    a       c       c       a 
7  -^  -  X  -  =  7  and  7  X  -  h-  -  =  7 ; 
b       d       d       b  b       d       d       b 

in  other  words,  that  multiplication  and  division,  as  defined  in 
§§  118,  119,  are  inverse  operations.     Compare  §  55. 

For,  by  §§  118,  119  and  §§  100,  111,  we  have 

a     c     c  _ad  c  _  adc  _a  dc  _a     a    c  .  c  _ac      c  _  acd  _n  rd  _a 
b      d    d      be   d      bed      b  cd      b'    b     d      d     bd      d      bdc      b  dc      b 

Hence  we  may  describe  the  kind  of  division  now  before  us 
as  the  inverse  of  multijMcation  and  say 

125  To  divide  a/b  Inj  c/d  is  to  find  a  number  ivhich  multiplied 
by  c/di  ivill  produce  a/b. 

By  introducing  fractions  into  our  number  system,  we  have 
made  it  possible  always  to  find  such  a  number,  except  when  the 
divisor  c/d  is  0. 


DIVISION   AND   FRACTIONS  37 

This  is  the  usual  meaning  of  division  in  arithmetic  and 
algebra.     It  is  the  generalization  of  exact  division,  §  93. 

Reducing  a  fraction  to  its  lowest  terms.     Irreducible  fractions.     126 
If  the  numerator  and  denominator  of  a  fraction  have  a  common 
factor,  we  can  remove  it  from  both  without  changing  the  value 
of  the  fraction. 

For  —  =  -,  since  am  ■  b  =  a-  bm,  §  106. 
bin      b 

When  all  such  common  factors  have  been  removed,  the  frac- 
tion is  said  to  be  in  its  lowest  terms,  or  to  be  irreducible. 

Theorem.     If  a/b  be  an  irreducible  fraction,  and  a'/b'  any     127 
other  fraction  which  is  equal  to  it,  then  a'  and  b'  are  equimul- 
tijjles  of  di  and  b  resjJectively. 

For  since  a' /b'  =  a/b.,  and  therefore  a'b  =  ab\  a  is  a  factor  of  a'b. 

But,  by  hypothesis,  a  has  no  factor  in  common  with  b.  Hence  a  must 
be  a  factor  of  a',  §  492,  1. 

We  therefore  have  a'  =  ma,  where  m  is  some  integer. 

But  substituting  ma  for  a'  in  a'b  =  ab',  we  have  mab  —  ab',  and  therefore 
b'  =  mb,  §  50. 

Corollary.     If  tivo  irreducible  fractions  are  equal,  their  numer-     128 
ators  must  be  equal,  and  also  their  denoniinato->'S. 


THE  USE   OF  FRACTIONS   IN   MEASUREMENT 

Fractional  lengths.     The  definition  of  length  given  in  §  81     129 
only  applies  to  such  line  segments  S  as  contain   the   unit 
segment  s  a  certain  number  of  times  exactly. 

But  even  if  S  does  not  contain  s  exactly,  it  may  still  be 
commensurable  with  s;  that  is,  it  may  contain  the  half  the  third, 
or  some  other  aliquot  part  of  s  exactly.  In  that  case  we  define 
its  length  as  follows  : 

If  a  given  line  segment  contains  th  e  hth  part  of  the  unit  segment     130 
a  times  exactly,  we  say  that  its  length  is  the  fraction  a/b. 


38  A   COLLEGE    ALGEBRA 

Thus,  if  S  contains  the  10th  part  of  s  exactly  7  times,  the  length  of  S 
(in  terms  of  s)  is  7  / 10. 

131  Note.  Observe  that  if  a/h  is  the  length  of  S  in  terms  of  s  according 
to  this  definition,  so  also  is  every  fraction  of  the  form  ma/mh. 

For  if  S  contains  the  6th  part  of  s  exactly  a  times,  it  will  contain  the 
j/i6th  part  of  s  exactlj'  ma  times. 

132  Fractions  are  useful  in  measurement  for  the  same  reason 
that  integers  are  useful :  namely,  hy  their  relative  positions 
in  the  rational  si/stem,  they  indicate  the  relative  sizes  of  the 
segments  whose  lengths  they  are. 

For  if  a/6  and  c/d  are  the  lengths  of  S  and  T  in  terms  of  s,  so  also 
are  ad/bd  and  bc/bd,  §  131 ;  that  is,  the  bdth  part  of  s  is  contained  in  S 
exactly  ad  times,  in  T  exactly  6c  times. 

Hence      S<,  =,  or  >T,  according  as     ad<,  =,  or  >6c, 
that  is,  S<,  =,  or  >T,  according  as  a/b<,  =  ,  or  >c/d.         §  106 

133  Note.  It  hardly  need  be  said  that  the  definition  of  length  here  given 
is  equivalent  to  the  definition  oi  fraction  given  in  elementary  arithmetic, 
and  that  greater  or  lesser  fractions  are  there  defined  as  fractions  v^hich 
correspond  to  greater  or  lesser  line  segments  or  other  magnitudes. 

134  Rational  numbers  pictured  by  points.  Fractions,  as  well  as 
integers,  may  be  pictured  by  points  on  an  indefinite  straight 
line,  §  85. 

_4  _3      _|_2  -1  0  1  2     ?  3  4 

>'  6         A  p 

Thus,  to  construct  a  point,  P,  which  will  picture  7/3  in  the  same  way 
that  A  pictures  1,  we  have  only  to  start  at  the  origin  0  and  lay  off  the 
third  part  of  the  unit  OA  seven  times  to  the  right. 

P',  the  corresponding  point  to  the  left  of  0,  is  the  picture  of  -  7/3. 

We  proceed  in  a  similar  manner  in  the  case  of  any  given  fraction, 
positive  or  negative. 

135  All  such  points  are  arranged  along  the  line  in  an  order 
corresponding  to  that  of  the  rationals  which  they  picture. 
With  this  in  mind  we  often  s])eak  of  one  rational  as  lying  to 
the  left  or  right  of  another  rational,  or  as  lying  between  two 
other  rationals. 


IRRATIONAL   NUMBERS  39 

IV.     IRRATIONAL   NUMBERS 

PRELIMINARY  CONSIDERATIONS 

Definitions.     The  product  aa  is  represented  by  a^,  read  "a     136 
square  "  ;  the  product  aaa,  by  a^  read  "  a  cube  "  ;  the  product 
aaa  ...  to  n  factors,  by  a",  read  "  the  nth  power  of  a." 

In  the  symbols  a^,  a^,  a",  the  numbers  2,  3,  n  are  called 
exponents;  a  itself  is  called  the  base. 

Finding  a^  from  a  is  called  squaring  a;  finding  a^,  cubing  b.) 
finding  a",  raising  a  to  ^A«3  /ith  power. 

The  operation  which  consists  in  raising  a  given  number  to 
a  given  power  is  also  called  involution. 

Roots  and  logarithms.     If,  as  we  are  supposing,  a  is  a  rational     137 
number,  and  n  a  positive  integer,  a"  is  also  a  rational  number. 
Call  this  number  b  ;  then 

a"  =  b. 

This  equation  suggests  two  new  problems  : 

First.     To  assign  values  to  n  and  b,  and  then  find  a. 

Second.     To  assign  values  to  a  and  b,  and  then  find  n. 

Thus,  (1)  let  n  =  2  and  6  =  9.     The  equation  then  becomes 

and  we  find  that  a  =  3  or  -  3 ;  for  both  32  =  9  and  (-  3)2  =  9. 
Again,  (2)  let  a  =  2  and  6  =  8.     The  equation  then  becomes 
2"  =  8, 
and  we  find  that  n  =  3 :  for  2^  =  8. 

When  a»  =  b,  138 

1.  a  is  called  the  nth  root  of  b,  and  is  expressed  in  terms 
of  n  and  b  by  the  symbol  v^,  the  simpler  symbol  V^,  read 
"  square  root  of  &,"  being  used  when  n  =  2. 

2.  n  is  called  the  logarithm  ofh  to  the  base  a,  and  is  expressed 
in  terms  of  a  and  b  by  the  symbol  log„  b. 


40  A   COLLEGE    ALGEBRA 

Thus,  since  3^  =  9  and  (—  3)-  =  9,  both  3  and  —  3  are  square  roots  of 
9,  and  both  may  be  written  V9 ;  but  see  §  139. 

Again,  2  is  the  logarithm  of  9  to  the  base  3 ;  that  is,  2  =  logs  9- 

139  Note.  Instead  of  representing  both  the  square  roots  of  9  by  the  symbol 
V9,  we  may  represent  the  positive  one,  3,  by  v9,  and  the  negative  one, 
—  3,  by  —  V9.  This  is  the  usual  method  of  representing  square  roots  in 
elementary  algebra,  and  we  shall  follow  it. 

140  Evolution  and  finding  logarithms.  The  operation  by  which 
V6  is  found,  when  n  and  b  are  given,  is  called  extracting  the 
nth  root  of  h,  or  evolution. 

The  operation  by  which  log^b  is  found,  when  a  and  b  are 
given,  is  called  Jinding  the  logarithm  o/b  to  the  base  a. 

Both  these  operations  are  inverses  of  involution,  §§  55,  124. 

141  Note.  The  reason  that  involution  has  two  inverses,  while  addition  and 
multiplication  each  has  but  one,  will  be  seen  by  comparing  the  three 
equations 

\.    a  -\-h  =  c.  2.    ab  =  c.  3.   a^  =  c. 

Since  a  +  b  =  b  -{-  a,  and  ab  =  ba,  the  problem :  Given  c  and  6  in  1  or 
2,  find  a,  is  of  the  same  kind  as  the  problem :  Given  c  and  a,  find  b. 

But  since  a*  is  not  equal  to  6",  the  problem  :  Given  c  and  b  in  3,  find 
a,  is  wholly  different  in  kind  from  the  problem  :  Given  c  and  a,  find  6. 

14J?  New  numbers  needed.  We  shall  subsequently  study  these 
new  operations  in  detail ;  for  in  algebra  they  are  second  in 
importance  to  the  four  fundamental  operations  only.  But  the 
point  which  now  concerns  us  is  this:  Tlie//  necessitate  further 
extensions  of  the  number  system. 

In  fact,  it  is  at  once  evident  that  vo  can  denote  a  rational 
number  in  exceptional  cases  only. 

Thus,  to  cite  the  simplest  of  illustrations,  neither  V^  nor  V2  can 
denote  a  rational  number.     For 

1.  Since  the  square  of  every  rational  number  is  positive,  no  rational 
exists  whose  square  is  —  1.  HeuCe  V—  1  cannot  denote  a  rational 
number. 

2.  No  rational  number  exists  who.se  siiuare  is  2.  For  clearly  2  is  not 
the  square  of  any  integer,  and  we  can  show  as  follows  that  it  is  not  the 
square  of  any  fraction. 


IRRATIONAL   NUMBERS  41 

Suppose  p/q  to  he  a  fraction  in  its  lowest  terms,  such  that 

(p/g)2  =  2,  or  p'^/q-^  =  2/1. 

But  since  p^/q~  is  in  its  lowest  terms,  §  492,  2,  it  would  follow  from 
\his,  §  128,  that  jj-  =  2,  wliich  is  impossible,  since  p  is  an  integer. 
Therefore  V2  cannot  denote  a  rational  number. 
It  can  be  shown  in  the  same  way  that  if  a/b  be  any  fraction  in  its 

lowest  terms,   ^a/b  cannot  denote  a  rational  number,  unless  both  a  and 
b  are  nth  powers  of  integers. 

We  are  to  make  good  this  deficiency  in  our  number  system 
by  creating  two  new  classes  of  numbers  :  the  irrational  num- 
bers, of  which  v2  is  one,  and  the  imaginary  mimbers,  of  which 
V—  1  is  one. 

We  shall  treat  the  irrational  numbers  in  the  present  chapter 
and  the  imaginary  numbers  in  the  chapter  which  follows. 

THE   ORDINAL   DEFINITION   OF   IRRATIONAL   NUMBERS 

In  the  present  chapter  the  letters  a,  b,  c,  and  so  on,  will 
denote  any  rational  numbers,  whether  positive  or  negative, 
integral  or  fractional. 

General  properties  of  the  rational  system.     The  rational  num-     143 
bers  constitute  a  system  which  has  the  following  properties  : 

1.  It  is  an  ordinal  system. 

2.  It  is  dense  ;  that  is,  between  every  two  unequal  numbers 
of  the  system,  a  and  b,  there  lie  still  other  numbers  of  the 
system. 

3.  The  sum,  difference,  product,  and  quotient  of  every  two 
numbers  of  the  system  are  themselves  numbers  of  the  system, 
the  quotient  of  any  number  by  0  excepted. 

By  the  definitions  which  follow,  we  shall  create  a  more 
extended  system  which  possesses  these  same  three  properties, 
and  which  includes  the  rational  system. 

Separations  of  the  first  kind.     1.    The  number  J  separates  the     144 
remaining  numbers  of  the  rational  system  into  two  classes : 


42  A   COLLEGE   ALGEBRA 

the  one  class  consisting  of  all  rationals  wliicli  jyrecede  (are  less 
than)  ^,  the  other  of  all  rationals  which  folloiv  (are  greater 
than)  I.  Let  us  name  these  two  classes  of  numbers  C^  and  C'2 
respectively. 

Cj I ^ 

In  the  figure,  the  half  line  to  the  left  of  the  point  \  contains  the  point- 
pictures  of  all  numbers  in  the  class  Ci,  and  the  half  line  to  the  right  the 
point-pictures  of  ail  numbers  in  the  class  C2,  §  134. 

From  §§  109,  111,  and  113,  it  immediately  follows  that 

1.  Each  number  in  C\  precedes  every  number  in  Co. 

2.  There  is  no  last  number  in  C\,  and  no  first  in  C\. 

Thus,  were  there  a  last  number  in  Ci,  there  would  be  numbers  between 
it  and  1/3,  §  113,  which  is  impossible  since,  by  hypothesis,  all  rationals 
less  than  1  /3  are  in  Ci. 

145  2.  Instead  of  thus  separating  the  rational  system  into  the 
three  parts  Cj,  \,  Co,  we  may  join  \  to  C'l,  so  forming  a  class 
C'l'  made  up  of  C\  and  \,  and  then  say: 

The  number  1  separates  the  entire  rational  system  into  two 
parts,  C'l'  and  C\,  such  that : 

1.  Each  number  in  C'/  precedes  every  number  in  Cg. 

2.  There  is  a  last  number  in  C/,  namely  \,  but  there  is  no 
first  number  in  (\. 

146  3.  Or  we  may  join  \  to  Co,  call  the  resulting  class  Co',  and 
then  say  : 

The  number  \  separates  the  entire  rational  system  into  two 
parts,  Ci  and  Cj',  such  that  : 

1.  Each  number  in  Ci  precedes  every  number  in  C2'. 

2.  There  is  no  last  number  in  Cj,  but  there  is  z.  first  number 
in  Co',  namely  \. 

It  is  evident  that  each  of  the  rational  numbers  defines 
similar  separations  of  the  rational  system. 

147  Conversely,  if  we  are  able,  iijany  way,  to  separate  the  entire 
rational   system   into  two  parts,  Bx  and  B^,  such  that  each 


IRRATIONAL    NUxMBERS  43 

number  in  B^  precedes  every  number  in  B2  and  that  there  is 
either  a  last  number  in  B^  or  a  first  in  B^,  the  separation  will 
serve  to  distinguish  this  last  or  first  number  from  all  other 
numbers  and,  in  that  sense,  to  define  it. 

Thus,  let  us  assign  the  negative  rationals  to  Bi  and  the  remaining 
rationals  to  B^-  There  is  then  no  last  number  in  Bi,  but  0  is  the  first 
number  in  B^.  And  zero  is  distinguished  from  all  other  numbers  when 
called  the  first  number  in  B2,  as  perfectly  as  by  the  symbol  0. 

Note.     Obviously  there  cannot  be  both  a  last  number  in  Bi  and  a  first      148 
in  B2.     For  there  must  then  be  rationals  between  these  two  numbers, 
§  113,  whereas,  by  hypothesis,  every  rational  belongs  either  to  Bi  or  to  B^. 

Separations  of  the  second  kind.     But  we  can  also,  in  various     149 
ways,  separate  the  eiitire  rational  system  into  a  part  A^  in 
which  there  is  no  last  number,  and  a  part  Ao  in  which  there  is 
no  first  number. 

Thus,  since  no  rational  exists  whose  square  is  2,  §  142, 
every  rational  is  either  one  whose  square  is  less  than  2,  or 
one  whose  square  is  greater  than  2. 

Let  A2  consist  of  all  positive  rationals  whose  squares  are 
greater  than  2,  and  let  Ai  consist  of  all  the  other  rational 
numbers.     Then 

1.  Each  number  in  A^^  precedes  every  number  in  A^. 

For  let  a\  be  any  number  in  ^1,  and  ai  any  number  in  An. 
Evidently  ai<a2,  if  ax  is  negative  or  0;  and  if  ai  is  positive,  ai2<a2^ 
and  therefore  a\  <  a^. 

2.  There  is  no  last  number  in  /li,  and  no  first  in  A^. 

For  when  any  positive  rational,  ai,  has  been  assigned  whose  square  is 
less  than  2,  we  can  always  find  a  greater  rational  whose  square  is  also 
less  than  2,  §  183,  2  (3)  ;  hence  no  number  can  be  assigned  which  is  the 
last  in  Ay.     Similarly  no  rational  can  be  assigned  which  is  the^irs^  in  A^. 

The  new  number  a  =  V2.     The  relation    between    the  two     150 
classes    of    numbers,   A^  and  A^,   is   therefore   jirecisely    the 
same  as  that  between  the  classes  C^  and  C2  in  the  separation 
corresponding  to  |-,  which  was  described  in  §  144. 


44  A   COLLEGE   ALGEBRA 

But  no  rational  number  exists  which  can  be  said  to  corre- 
spond to  the  separation  Ai,  A^,  or  to  be  defined  by  it. 

For  since  every  rational  belongs  either  to  A^  or  to  A  2,  no 
rational  exists  which  lies  between  A^  and  A  2,  as  i  lies  between- 
Ci  and  Cg. 

And  since  there  is  no  last  number  in  A^  and  no  first  in 
A  2,  no  rational  exists  which  corresponds  to  this  separation  as 
^  corresponds  to  the  separation  Cj',  Cg  of  §  145,  or  to  the 
separation  C\,  C^'  of  §  146.     (Compare  §  147.) 

Hence  this  separation  Ai,  .1 2^ creates  a  place  for  a  netv  ordi- 
nal number,  namely,  a  number  which  shall  follow  all  numbers 
in  A I  and  precede  all  iiv  A  2. 

We  invent  such  a  number.  For  the  present  we  may  repre- 
sent it  by  the  letter  a;  later,  when  multiplication  has  been 
defined  for  a,  we  shall  find  that  a^  =  2,  and  we  can  then 
replace  a  by  the  more  significant  symbol  V2,  §  182. 

151  We  then  define  this  new  number  a  as  that  number  which  lies 
between  all  positive  rationals  whose  squares  are  less  than  2 
and  all  whose  squares  are  greater  than  2. 

We  may  also  express  this  definition  by  the  formula 
«!  <  a  <  as 
where  a^  and  02  denote  any  numbers  whatsoever  in  Ax  and  A2 
respectively,  and,  as  heretofore,  <  means  "  precedes." 

152  Note.  Observe  that  this  definition  is  of  the  same  kind  as  the  defini- 
tions of  the  negative  and  fractional  numbers  given  in  §§  56,  110.  Like 
these  numbers,  a  is  a  symbol  defined  by  its  position  in  an  ordinal  system  of 
symbols  which  includes  the  natural  numbers.  It  therefore  has  precisely 
the  same  right  as  they  to  be  called  a  number. 

Our  reason  for  inventing  this  and  similar  numbers  is  also  the  same  as 
our  reason  for  inventing  negative  numbers  and  fractions.  They  serve  a 
useful  purpose  in  the  study  of  relations  among  the  numbers  which  we 
already  possess,  and  among  things  in  the  world  about  us.* 

*  We  may  add  that  there  would  be  no  objection  from  an  ordinal  point  of 
view  to  our  inventing  more  than  one  number  to  correspond  to  the  separation 
A■^,  A^,  say  two  numbers,  a  and  b,  defined  ordinally  by  the  formula  «,  <a<b<  a^. 

But  there  are  objections  of  another  kind  to  our  inventing  more  than  one 
such  number.    See  page  67,  footnote  (3). 


•IRRATIONAL   NUMBERS  45 

The  irrational  numbers  in  general.     The  real  system.     The     153 

particular  separation  of  the  rational  system  which  we  have 
been  considering  is  but  one  of  an  infinite  number  of  possible 
separations  of  a  similar  character. 

For  every  such  separation  we  invent  a  new  number,  defining 
it  ordinally  with  respect  to  the  numbers  of  the  rational  system 
precisely  as  we  have  defined  the  number  a  =  V2  in  §  151. 

To  distinguish  these  new  numbers  from  the  rational  num- 
bers, we  call  them  irrational  numbers,  or  simply  irrationals. 

Again,  to  distinguish  the  rational  and  irrational  numbers 
alike  from  the  imaginary  numbers,  which  we  have  yet  to 
consider,  we  call  them  ro/xl  numbers. 

Finally,  we  call  the  system  which  consists  of  all  the  rational 
and  irrational  numbers  the  system  of  real  numbers,  or  the  real 


Hence,  using  a  to  denote  any  irrational  number,  we  have  the 
following  general  definition  of  such  a  number  : 

An  irrational  number,  a,  is  defined  ivhenever  a  latv  is  stated  154 
which  will  assign  every  given  rational  to  one,  and  but  one,  of 
two  classes,  Aj,  Ag,  such  that  (1)  each  number  in  A^  precedes 
every  number  in  A«  and  (2)  there  is  no  last  number  in  Ai 
and  no  first  number  it}  Aa  ;  the  defiiiition  of  a  then  being:  it 
IS  the  one  number  ivhieh  lies  between  all  numbers  in  A^  and 
all  in  Ag. 

It  is  here  implied  that  there  are  numhers  in  both  the  classes  Ai  and 
Ai;  also  that  Ai  and  ^2  together  comprise  the  entire  rational  system. 

An  irrational  number,  a,  is  said  to  be  negative  or  positive     155 
according  as  it  precedes  or  follows  0. 

The  real  system  is  an  ordinal  system :  that  is,  the  numbers  156 
which  constitute  it  are  arranged  in  a  definite  and  known  order, 
§  17.  For  the  definition  of  each  irrational  indicates  how  it 
lies  with  respect  to  every  rational:  and  from  the  definitions 
of  any  two  given  irrationals  we  can  at  once  infer  how  they  lie 
with  respect  to  one  another. 


46  A   COLLEGE   ALGEBRA 

Thus  let  a  and  b  denote  any  two  given  irrationals ;  then 
1.  If  every  rational  which  precedes  a  also  precedes  b,  and 
every  rational  which  follows  a  also  follows  b,  the  numbers  a 
and  b  occupy  the  same  position  relative  to  the  numbers  of  the 
rational  system.  By  our  definition  of  an  irrational  number, 
therefore,  §  154,  a  and  b  denote  one  and  the  same  number. 
We  indicate  this  by  the  formula : 


2.  If  among  the  rationals  which  follow  a  there  are  some 
which  precede  b,  then  a  itself  must  precede  b  ^or  b  follow  a). 
We  indicate  this  by  the  formula : 

a  <  b  or  b  >  a. 

3.  If  among  the  rationals  which  precede  a  there  are  some 
which  follow  b,  then  a  itself  must  follow  b  (or  b  precede  a). 
We  indicate  this  by  the  formula : 

a  >  b  or  b  <  a. 

157  It  thus  appears  that  when  any  two  different  real  numbers 
are  given,  we  can  at  once  infer  which  precedes  and  which 
follows;  also,  that  we  may  always  draw  the  following  con- 
clusions with  respect  to  three  given  real  numbers,  a,  b,  c : 

If  a  =  b,  and  b  =  c,  then  a  =  c. 
If  a  <  b,  and  b  <  c,  then  a  <  c. 
If  a  =  b,  and  b  <  c,  then  a  <  c. 

158  The  real  system  is  dense.  For  there  are  rational  numbers 
not  only  between  any  two  unequal  rationals,  §  113,  but  also 
between  any  two  unequal  irrational  numbers,  and  between  any 
two  numbers  one  of  which  is  rational  and  the  other  irrational, 
§  15  a 

159  The  real  system  is  continuous.  The  real  system,  therefore, 
possesses  the  first  and  second  of  the  properties  of  the  rational 


IRRATIONAL   NUMBERS  47 

system  enumerated  in  §  143.     But  it  possesses  an  additional 
property  not  belonging  to  the  rational  system,  namely  : 

If  the  entire  real  system  he  separated  into  tivo  parts,  R^ 
and  R25  such  that  each  number  in  E-i  precedes  every  number 
in  Rjj  there  is  eitJier  a  last  nutnber  in  Ri  or  a,  Jirst  in  Rj,  but 
not  both. 

For  in  separating  the  real  system  into  the  parts  i?i  and  E^,  we  separate 
the  rational  system  into  two  parts,  A^  and  A^,  the  one  part  consisting  of 
all  the  rationals  in  i?i,  the  other  of  all  the  rationals  in  i?2. 

Every  rational  belongs  either  to  ^1  or  to  A^,  and  each  rational  in  Ai 
precedes  every  rational  in  A^. 

Let  a  be  the  umnber  which  the  separation  A^,  An  defines,  §§  147,  154. 

Then  either  a  is  a  rational  —  namely  the  last  number  in  ^  1  or  the  first 
in  A2,  §  147,  —or,  if  there  be  no  last  number  in  Ai  and  no  first  in  A^, 
a  is  an  irrational  lying  between  Ai  and  An,  §  154. 

1.  If  a  is  the  last  number  in  Ai,  it  is  also  the  last  in  jRi;  for  were 
there  any  number  in  Bi  after  a,  there  would  be  rationals  between  it  and  a, 
that  is,  rationals  in  Ai  after  a,  which  is  impossible. 

2.  Similarly,  if  a  is  the  first  number  in  A^.,  it  is  also  the  first  in  R^. 

3.  If  a  is  irrational,  it  must,  by  hypothesis,  belong  either  to  Ei  or  to 
i?2-  If  a  belongs  to  Ei,  it  is  the  last  number  in  Ri ;  for  were  there  any 
number  in  i?i  after  a,  there  would  be  rationals  between  it  and  a,  §  158, 
that  is,  rationals  in  Ai  after  a,  which  is  impossible.  And,  in  like  manner, 
if  a  belongs  to  E2,  it  is  the  first  number  in  E^. 

Finally,  there  cannot  be  both  a  last  number  in  Ei  and  a  first  in  E2, 
since  there  would  be  rationals  between  these  two  numbers,  §  158,  that  is, 
rationals  belonging  neither  to  Ai  nor  to  A^  ;  which  is  impossible. 

To  indicate  that  the  real  system  is  dense  and  at  the  same 
time  possesses  the  property  just  described,  we  say  that  it  is 
conti7iiioiis. 

Theorem.      A  real  number,  a,  either  rational  or  irtiational,  is     160 
defined  whenever  a  law  is  stated  by  means  of  ivhich  the  entire 
real  system  m,ay  be  separated  into  tivo  parts,  Rj,  R2,  such  that 
each  number  in  Ri  precedes  every  number  in  R2  ;  this  number,  a, 
tlien  being  either  the  last  numher  in  Rj  or  the  first  in  Rj.' 

This  is  an  immediate  consequence  of  §§  147,  159. 


48  A   COLLEGE    ALGEBRA 


APPROXIMATE   VALUES   OF   IRRATIONALS 

161  Given  any  irrational  number,  a,  defined  as  in   §  154.     By 

the  method  illustrated  below  we  can  find  a  pair  of  rationals, 
the  one  less  and  the  other  greater  than  a,  which  differ  from 
each  other  as  little  as  we  please.  Such  rationals  are  called 
approximate  values  of  a. 

Let  a  be  the  irrational,  V2,  which  lies  between  all  positive  rationals 
whose  squares  are  less,  and  all  whose  squares  are  greater  than  2. 

1.  We  may  find  between  what  pair  of  consecutive  integers  a  lies  by 
computing  the  squares  of  1,  2,  3,  ■  •  ■  successively,  until  we  reach  one 
which  is  greater  than  2. 

We  see  at  once  that  1^  <  2  and  22  >  2. 
Hence  a  lies  between  1  and  2,  or  1  <a<2. 

2.  We  may  then  find  between  what  pair  of  consecutive  tenths  a.\ies  by 
computing  Ll^,  1.2^,  •  •  •  successively,  until  we  reach  one  which  is  greater 
than  2. 

We  thus  obtain  1.42<2and  1.52>2;  for  1.42=  i.ge^  1.52=:  2.25. 
Hence  a  lies  between  1.4  and  1.5,  or  1.4<a<1.5. 

3.  By  a  similar  procedure  we  find,  successively, 

1.41<a<1.42,     1.414<a<1.4l5,     and  so  on  without  end. 

4.  Let  ai  denote  the  nth  number  in  the  sequence  1.4,  1.41,  1.414,  •  •  • 
thus  obtained,  and  Uo  the  nth  number  in  the  sequence  1.5,  1.42,  1.415,  •  •  • . 

Then  ai<a<a2  and  02  -  ai  =  1/10", 

and  by  choosing  n  great  enough  we  can  make  1/10"  less  than  any  positive 
number,  as  5,  we  may  choose  to  assign,  however  small. 

5.  We  call  1.4,  1.41,  1.414  the  approximate  values  of  a  =  V2  to  the 
first,  second,  third  place  of  decimals;  and  so  on. 

Evidently  the  process  thus  illustrated  may  be  applied 
to  any  given  irrational  number,  a;  for  all  that  the  process 
requires  is  a  test  for  determining  whether  certain  rationals 
are  less  or  greater  than  a,  and  the  definition  of  a,  §  154,  will 
always  supply  such  a  test.  We  therefore  have  the  following 
theorem : 


IRRATIONAL   NUMBERS  49 

Let  a  denote  any  given  irrational  number.     If  any  positive     162 
number,  as  8,  be  assigned,  it  matters  not  hoiv  small,   we  can 
always  find  two  rationals,  ai,  aj,  such  that 

ai  <  a  <  a2  and  a.i  —  ai  <  8. 

Evidently  this  theorem  is  true  of  rationals  also. 

Thus,  if  a  denote  a  given  rational  number,  and  ai  =  a  —  1/10", 
«.,  =  a  +  1/10",  we  have  ai<a<a2,  and  we  can  make  a^  —  ai  =  2/10"  as 
small  as  we  please  by  choosing  n  sufficiently  great. 


ADDITION,    SUBTRACTION,    MULTIPLICATION,    DIVISION 

It  remains  to  give  the  real  system  the  third  of  the  proper- 
ties of  the  rational  system  enumerated  in  §  143.  For  this  we 
shall  require  the  following  theorem  :  ^ 

Theorem.     Let  Ai  and  Aj  be  two  classes  of  rationals  such  that     163 

1.  Each  number  in  A^  is  less  than  every  number  in  A^, 

2.  There  is  no  last  number  in  A^  and  no  first  hi  Aj, 

3.  For  every  positive  number,  8,  that  may  be  assigned,  it 
matters  not  how  small,  ice  can  find  in  A^  a  number  a^,  and  in  A^ 
a  number  ao,  such  that 

3-2  —  ^1  <  8. 

We  may  then  conclude  that  between  A^  and  Aj  tltere  lies  one 
number  and  but  one. 

That  there  is  at  least  one  such  number  follows  from  1  and  2,  by  §  154. 

That  there  cannot  he  more  than  one  such  number  follows  from  3. 

For  suppose  that  between  every  oi  in  Ai  and  every  az  in  A^  there  were 
the  two  rationals  d  and  d',  as  indicated  in  the  figure  : 


Then  for  every  ai,  a2  we  should  have 

a2>d',  and   -«i>-d,  §§73,121 

and  therefore  as  -  ai  >d'  -  d,  §§  39,  121 

which  is  impossible,  since  it  contradicts  3. 


50  A   COLLEGE   ALGEBRA 

Nor  can  there  be  two  numbers,  one  or  both  of  which  are  irrational^ 
Ij'ing  between  every  ai  and  a^ ;  for  between  these  two  numbers  there 
would  be  two  rationals  also  lying  between  every  ai  and  as,  §  158,  which 
we  have  just  shown  to  be  impossible. 

164  Note.  This  theorem  differs  from  the  definition  of  an  irrational  number, 
§  154,  in  that  it  is  not  here  a  part  of  the  hypothesis  that  every  rational  lies 
either  in  Ai  or  in  A2. 

165  Addition.  Let  a  and  b  denote  any  two  ffiven  real  numbers, 
rational  or  irrational,  and  let  aj,  a2,  h^  h^,  denote  any  rationals 
whatsoever  such  that 

«!  <  a  <  02  and  ij  <  b  <  b^-  (1) 

Observe  that  there  is  no  last  number  of  the  kind  denoted 

by  rti  or  ^1,  and  no  first  number  of  the  kind  denoted  by  a^  or 

Z'a ;  and  that  if  any  positive  number,  as  8,  be  assigned,  it  matters 

not  how  small,  we  can  alw^ays  choose  «i,  a^,  and  h^,  h»,  §  162, 

so  that  both 

a,  -  «i  <  8  and  h.,  -  h,  <  8.  (2) 

When  both  a  and  b  are  rational,  say  a  =  a  and  b  =  ^,  we  can 
find  their  sum,  a  +  /3,  by  the  rule  of  §  116 ;  and  it  follows 
from  (1),  by  §  121,  that 

«!  +  ^1  <  «  +  /3  <  «2  +  ^^2- 

Moreover,  whether  a  and  b  are  rational  or  not,  it  follows 
from  (1),  by  §  121,  that 

«!   +  Z>i  <   «2  +  K  (3) 

These  considerations  lead  us  to  define  the  sum  of  a  and  b, 
when  one  or  both  are  irrational,  as  follow^s  : 

166  The  Slim  of  &  and  b,  vriften  a  +  b,  is  to  mean  that  number 
'which  lies  between  all  the  numbers  a,  +  bi  and  all  the  numbers 
a2  +  bg.     In  other  words,  it  is  the  number  defined  by  the  formula 

ai  +  bi  <  a  4-  b  <  a.,  +  b.,, 

where  aj,  aj,  bj,  bo  denote  any  rationals  whatsoever  such  that 

ai  <  a  <  ao  and  bi  <  b  <  bj. 


IRRATIONAL   NUMBERS  61 

To  justify  this  definition  we  must  show  that  there  is  one  and  but  one 
such  number  a  +  b.     This  follows  from  §  163  ;  for 

1.  Each  ai  +  6i  is  less  than  every  a<i  +  6o. 

2.  There  is  no  last  ai  +  6i  and  no  first  a-z  +  62. 

Thus,  Ui  +  bi  cannot  be  the  last  ai  +  61 ;  for  since  there  is  no  last  ai 
and  no  last  61,  we  can  choose  a\  and  bi  so  that  ai>ai  and  6i>6i',  and 
therefore  ai  +  61  >  a/  +  &i'. 

3.  If  any  positive  rational,  5,  be  assigned,  we  can  choose  ai,  aa,  &i,  62, 

so  that 

a2-ai<5/2  and  62 -6i<5/2,  §102 

and  therefore  {ao  +  62)  -  (ai  +  bi)  <8.  §  121 

Definition  of  —  a.     Let  a,  a^,  a^  have  the  same  meanings  as  in     167 
§  165.     Considerations  like  those  in  §  165  lead  us,  when  a  is 
irrational,  to  define  —  a  as  follows  : 

The  symbol  —  a  is  to  mean  the  number  defined  by  the  formula     168 

—  aa  <  —  a  <  —  ai, 

where  ai,  a2  denote  any  i-ationals  whatsoever  sxich  that 

ai  <  a  <  a2. 

It   follows   from  §  163  that  there  is  one  and  but  one  such  number 

—  a  ;  for 

1.  Each  —  ai  is  less  than  every  —  ai,  since  ai<ao.  §§  73,  111 

2.  There  is  no  last  —  ai  and  no  first  —  a\.     Thus,  were  there  a  last 

—  02,  there  would  be  a  first  a-i ;  but  no  such  number  exists. 

3.  We  can  always  choose  ai,  a2,  so  that 

-ai-(-a2)  =  a2-ai<5.  §162 

Subtraction.      The  result   of  subtracthuj  b   from  a,   written     169 
a  —  b,  is  to  mean  the  number  a  +  (—  b) ;  that  is, 

a-b  =  a+(-b). 
The  meaning  of  a  +  (—  b)  itself  is  known  from  §§  166,  168. 
It  follows  from  §§  166, 168  that  a  —  b  may  also  be  defined  by  the  formula 
ai  —  62  <  a  —  b  <  ao  —  61, 
where  ai,  a^^  b\,  bo,  denote  any  rationals  whatsoever  such  that 
ai<a<a2  and  6i<b<62- 


52  A   COLLEGE   ALGEBRA 

170  Multiplication,  both  factors  positive.     Let  a  and  b  be  any  two 

given  2)osi.tive  numbers,  and  a^,  a^,  h^,  ho,  any  jjositive  rationals 
whatsoever  such  that 

tti  <  a  <  fl.,  and  bi<h  <  h^.  (1) 

When  a  and  b  are  rational,  say  a  =  a,  b  :=  /3,  it  follows  from 
(1),  by  §  121,  that 

and  in  every  case  it  follows  that 

a  A  <  a^ho.  (2) 

We  are  therefore  led,  when  one  or  both  of  the  numbers  a,  b 
are  irrational,  to  define  their  product  thus  : 

171  The  product  of  two  positive  mimbers  a  and  b,  ivritten  ab,  is  to 

mean  that  number  \ohich  lies  between  all  the  numbers  ajbi  and 

all  the  numbers  aobo.     In  other  tvords,  ab  is  the  number  defined 

by  the  formula 

aibi  <  ab  <  aob.,, 

where  aj,  ao,  bj,  bj  denote   any  ])ositire  rationals  ^vhatsoever 

such  that 

ai  <  a  <  a2  and  bi  <  b  <  bo. 

It  follows  from  §  103  that   there  is   one  and  hut  one  such  number 
ab;  for 

1.  Each  a^hx  is  less  than  every  a^hi- 

2.  There  is  no  last  ciihi  and  no  first  a^ho.     (Compare  proof,  §  166,  2.) 

3.  Any  positive  5  being  given,  we  can  choose  ai,  a-z,  6i,  h2,  so  that 

aJtio  —  aihi  <  5. 
For  a.262  —  «i'Ji  =  «2  ('>:  -  ^1)  +  &i  {do  —  tti), 

and  we  can  choose  ai,  a-i,  h\,  h-2,  §  16'2,  so  that 

&2  -  61  <  5/2  ffl2  and  no  -  ffj  <  5/2  ^^i,  (1) 

and  therefore  ao  (62  —  &i)  +  ^^i  {(lo  -  ai)  <  S.  (2) 

■    We  may  make  such  a  choice  of  ai,  ao,  /*i,  ft.,,  as  follows : 

First  take  any  particular  number  of  the  kind  bo,  as  62',  and  then  choose 
Oi)  Ozj  so  that 

as- ai<5/2  62'.  (3) 


IRRATIONAL   NUMBERS  53 

Next,  using  the  ao  thus  found,  choose  61,  62,  so  that 

62  —  f'i<5/2a2,  as  in  (1). 

Since  61  <  h-z  and  therefore  5/2  62'  <  5/2  61,  it  follows  from  (3)  that 

a-z  -  ai<5/2  6i,  as  in  (1). 

Multiplication,  one  or  both  factors  negative  or  0.     Let  a  and  b     172 

denote  any  two  given  positive  numbers.     Then 

1.  a  (—  b)  and  (—  a)  b  are  to  mean  —  ab. 

2.  (—  a)  (—  b)  is  to  mean  ab. 

3.  a  ■  0  and  0  •  a  are  to  mean  0. 

Definition  of  1/a.     Let  a  be  any  given  positive  number,  and     173 

cii,  Go,  any  positive  rationals  vk^liatsoever  such  that 

rti  <  a  <  og- 

Considerations  like  those  of  §  165  lead  us,  when  a  is  irra- 
tional, to  define  1/a  as  follows  : 

The  symbol  1/a  is  to  viean  the  number  defined  by  the  formula     174 

l/fl2<  l/a<  1/ai, 

ivhere  ai,  ao  denote  any  2^ositire  rationals  whatsoever  such  that 

«!  <  a  <  ao. 

It  follows  from  §  103  that  there  is  one  and  but  one  such  number 
1/a;  for 

1.  Each  l/tto  is  less  than  every  l/oi,  §  106. 

2.  Tliere  is  no  last  l/a2  and  no  first  1/ai.     (Compare  proof,  §  168,  2.) 

3.  Any  positive  5  being  given,  we  can  choose  ai,  ao  so  that 

1/ai-  l/a2<5. 

For  1/ffi -l/«2<5,  if  a2-ai<5-airt2.  §§106,117 

But  if  ai  denote  any  particular  number  of  the  kind  «!,  we  can  choose 
ai,  ao  so  that  ai  >  ai  and  an  —  ai<  5ai'-,  and  therefore  <  5aiOo. 

Definition   of  l/(— a).     Let   a   denote    any    given    positive     175 
number.      Then  l/(— a)   is  to  mean  —1/a. 


54  A   COLLEGE   ALGEBRA 

176  Division.  The  quotient  of  a  by  b  (b  not  0)  is  to  mean  the 
number  a  •  1  /b,  that  is, 

a  1 

b^^b- 

The  meaning  of  a  •  1/b  itself  is  known  from  tlie  preceding  definitions. 
When  a  and  b  are  positive,  it  follows  from  §§  171,  174  that  we  may 
also  define  a/b  by  the  formula 

ai/62<a/b<  02/61, 

where  ai,  02,  61,  62  denote  any  positive  rationals  whatsoever  such  that 

ai  <  a  <  a2  and  61  <  b  <  62. 

177  The  commutative,  associative,  and  distributive  laws.  The  oper- 
ations just  defined  are  extensions  of  the  corresponding  oper- 
ations for  rational  numbers.  Subtraction  continues  to  be  the 
inverse  of  addition,  and  multiplication  of  division.  Finally, 
addition  and  multiplication  continue  to  conform  to  the  commuta- 
tive, associative,  and  distributive  laws. 

Thus,  if  a,  b,  and  c  are  any  three  positive  numbers  defined,  as  in 
§  170,  by  the  formulas 

tti  <  a  <  a2,        61  <  b  <  62,        Ci  <  c  <  C2, 

we  have  a  (b  +  c)  =  ab  +  ac 

For  by  §§  166,  171,  a  (b  +  c)  and  ab  +  ac  are  defined  by  the  formulas 

ai(6i  +  Ci)<a(b  +  c)<a2(&o-|-C2),  (1) 

ffli^i  +  aiCi  <  ab  +  ac  <  0262  +  aiCi-  (2) 

And  since  ax  (61  +  Ci)  =  a^bi  +  aiCi  and  a^  {ho  +  co)  =  aJb^  +  a^Co,  §  120, 
the  numbers  defined  by  (1)  and  (2)  are  the  same. 

178  The  rules  of  equality  and  inequality.  These  also  hold  good 
for  sums  and  products  as  just  defined,  namely  : 

According  as  a<,  =,  or  >  b, 

so  is  a  -f  c  <,  =,  or   >  b  +  c; 

also,  ac  <,  =,  or   >  be,                        if  c  >  0, 

but  ac  >,  =,  or  <  be,                       if  c  <  0. 


IRRATIONAL   NUMBERS  55 

Thus,  if  a  <  b,  then  a  +  c  <  b  +  c. 

Tor  let  d  and  d  +  a  he  any  two  rationals  between  a  and  b,  and  choose 
Ci  so  that  Ci  <  c  <  Ci  +  a. 

Then,  since  a<d  and  c<Ci  +  a,  we  have  a  +  c<d  +  Ci  +  a,  (1) 

and  since  d  +  a<h  and  Ci < c,  we  have  d  +  cr  +  Ci < b  +  c,  §  166.  (2) 

But  from  (1)  and  (2)  it  follows,  §  157,  that  a  +  c<b  +  c. 

The  proof  that,  if  a<b  and  c>0,  then  ac<bc,  is  similar. 
But  in  this  case  we  choose  Ci  so  that  Ci<c<Ci(l  +  cx/d). 

From  these  rules  it  follows,  as  in  §  39,  that  if  a  <  b  and     179 
c  <  d,  then  a  +  c  <  b  +  d,  and  so  on ;  also,  as  in  §  50,  when 
a,  b,  c,  d  are  positive,  that  if  a  <  b  and  c  <  d,  then  ac  <  bd, 
and  so  on. 

On  approximate  values.     1.    Having  now  defined  subtraction     180 
for  irrational  numbers,  §  169,  we  can  state  the  theorem  of 
§  162  as  follows  : 

When  any  irrational  a  is  given,  and  any  positive  rational  S 
is  assigned,  hoivever  small,  we  can  always  find  ratiojials,  aj  and 
a2,  which  will  differ  from  a  by  less  than  S. 

For,  by  §  162,  we  can  find  ai  and  a^  such  that  ai  <  a  <  02  and  a^  —  ai<  5. 
But  from  a<rt2  it  follows,  §  178,  that  a  —  ai<a2  —  ai,  and  therefore 
that  a  -  ai  <  5. 

In  like  manner,  since  —  a  <  —  ai,  we  prove  that  ao  —  a  <  5. 
Thus,  §  161,  we  have  V2  -  1.41  <. 01  and  1.42  -  V2<.01. 

We  say  of  such  an  a^  or  a^  that  it  represents  a  with  an 
error  not  exceeding  S. 

2.  In  practical  reckoning  we  employ  approximate  values  of 
irrational  numbers  more  frequently  than  the  numbers  them- 
selves. If  «!  and  bi  are  approximate  values  of  a  and  b  respec- 
tively, then  ai  +  b^  will  be  an  approximate  value  of  the  sum 
a  +  b.  But  to  insure  that  the  error  of  «i  +  ^^i  shall  not  exceed 
8,  we  must  ordinarily  choose  a^  and  b^  so  that  their  respective 
errors  shall  not  exceed  8/2.  This  follows  from  the  proof  in 
§  166.  Similar  rules  for  finding  approximate  values  of  a  —  b, 
ab,  and  a/b  with  errors  not  exceeding  8,  may  be  derived  from 
the  proofs  in  §§  168,  171,  174. 


56  A   COLLEGE    ALGEBRA 

INVOLUTION   AND   EVOLUTION 

181  Powers.  In  the  case  of  irrational  numbers,  as  in  that  of 
rationals,  we  represent  the  products  aa,  aaa,  •  ■  •,  by  a-,  a^,  •  •  •. 

182  Roots.  The  mth  root  of  any  given  positive  number  b,  written 
'■Vb,  is  to  mean  that  positive  number  whose  mt\\  power  is  b ; 
that  is,  Vb  is  to  denote  the  positive  number  which  is  defined 
by  the  formula  ( Vb)'"  =  b. 

To  justify  this  definition,  we  must  show  that  one,  and  but 
one,  such  number  as  it  implies  actually  exists.  We  accomplish 
this  as  follows  : 

183  Theorem.  The  real  sijstem  contains  the  rath  root  of  every  posi- 
tive real  number  b. 

1.  If  b  is  the  ??ith  power  of  a  rational  number,  the  truth  of 
the  theorem  is  obvious. 

Thus,  if  b  =  8/27  =  (2/3)3,  then  Vb  =  2/3. 

2.  If  b  is  not  the  int\\  power  of  a  rational,  its  ?»th  root  is 
that  real  number  a  which  lies  between  all  positive  rationals, 
«!,  whose  mth  powers  are  less  than  b  and  all  positive  rationals, 
agj  whose  mth  powers  are  greater  than  b.     Compare  §  151. 

It  follows  from  §  1.54  that  there  is  one,  and  but  one,  such  number  a, 
since  (1)  every  positive  rational  is  either  an  ai  or  an  02,  (2)  each  a\  is  less 
than  every  a..,  and  (3)  there  is  no  last  a\  and  no  first  rt2. 

We  may  prove  (3)  as  follows  : 

If  there  be  a  last  ai,  call  it  p.  Then  since  ;)"'<b,  there  are  rationals 
between  p'"  and  b.  Let  one  of  them  be  p™  +  5.  We  have  only  to  show 
that  we  can  find  a  rational  q>p  such  that  7"'  <  p"'  +  5,  or  (/'"  —  p"'  <  5  ; 
for  we  shall  then  have  p'"  <  7'"  <  b,  so  that  p  is  not  the  last  rti. 

But  (/'"  -  p"'  =  {q-  p)  (7'"-'  +  r/'-'-^p  +  •  •  ■  +  qp'"-'  +  p"—')       §'308 

<(q  —  p)ma2.'"'~^,  if  02'  be  any  particular  a^, 

<5,  if  q  -p  +  d/viuo' "'-''■. 

We  ean  show  in  a  similar  manner  that  there  is  no  first  03. 

This  established,  it  may  readily  be  proved  that  a  =  v  b. 


IRRATIONAL   NUMBERS  57 

For,  since  ai<a<a2>  we  have  ai™ < a™ < a2'".  §§  171,  181 

But  b  is  the  only  number  between  every  a^"'  and  every  a^'". 
Hence  a"'  =  b,  tliat  Is,  a  =  A^b. 

Rules  of  equality  and  inequality.      Let  a   and   b   denote  any     184 
positive  real  numbers,  and  7u  any  positive  integer.     Then 

According  as  a  <,  =,  ov  >  b, 

so  is  a"'  <,  =,  or   >  b"',  (1) 

and  Va<,=^,  or   >  V^.  (2) 

We  may  prove  (1)  by  repeated  use  of  §  179. 

Thus,  if  a<b,  then  a-  a<b  -b,  that  is,  a^  <  b- ;  and  so  on. 

We  derive  (2)  from  (1).      Thus,  if  a  =  b,  then  v a  =  v^;   for  were 
Va  <  or  >  Vb,  we  should  have  a  <  or  >  b. 

Rules  of  exponents.     Let  a  and  b  denote  any  two  real  num-     185 
bers,  and  rn  and  7i  any  two  positive  integers.     Then 

1.    a'"  •  a"  =  a'"  + ".        2.    («'")"  =  «'"".        3.     (ab)"' =  a"'b"\ 
Thus,  a^  ■  a^  =  aaa  ■  aa      =  aaaaa  =  a^  —  a^  +  -  §177 

(a2)3  =  o2.a2.a2  =n2  +  2  +  2        =cfi-5  by  1 

{abf  =  abab-ab  =  aaa  ■  bbb       =a^.b^  §  177 

And  similarly  for  any  other  positive  integral  values  of  m  and  n. 

A  theorem  regarding  roots.     Let  a  and  b  denote  any  positive     186 
real  numbers,  and  7n  any  positive  integer.     Then 

Vrt  V6  =  VaZ. 

For  ( Va  •  'v^)'«  =  ( v'a)™  •  (v^)-"  =  ab     §§  182,  185,  3 

and  (\^b)'»  =  ab.  §  182 

Hence  (Va  •  \/6)"'  =  (v'^)"' 

and  therefore  'Va-"Vb=  Vab.  §  184,  (2) 


58  A   COLLEGE   ALGEBRA 

VARIABLES   AND  LIMITS 

187  Variables.     We  say  that  a  never-ending  sequence  of  num 

bers,  such  as 

a  I,  ito,  a  3,  •  •  • ,  «„,  •  •  • , 

is  (/ive7i  or  known,  if  the  value  of  every  particular  term  «„  is 
knowTi,  or  can  be  computed,  when  the  index  n  which  shows 
its  position  in  the  sequence  is  given. 

We  often  have  occasion  to  consider  variables  which  are 
supposed  to  be  running  through  such  given  but  never-ending 
sequences  of  values. 

Thus,  I,  f,  |,  •  •  •,  -^,  •  •  •  is  such  a  given  never-endinfj  sequence,  and 
X  is  such  a  variable  if  we  suppose  it  to  be  running  through  this  sequence, 
that  is,  to  be  talking  successively  the  values  |,  f ,  |,  •  •  •  • 

188  Limits.  As  x  runs  through  the  sequence  i,  §,  |,  •••,  it  con- 
tinually approaches  the  value  1,  and  in  such  a  manner  that  if 
we  assign  any  positive  number,  as  8,  it  matters  not  how  small, 
the  difference  1  —  x  will  ult  ately  become  and  remain  less 
than  the  number  so  assigned.  Thus,  after  x  reaches  the  100th 
term  of  the  sequence,  1  —  x  will  remain  less  than  .01. 

We  express  all  this  by  saying  that,  as  x  runs  through 
the  sequence  \,  §,  |,  ■  •  •,  it  approaches  1  as  limit.  And  in 
general 

189  A  variable  x,  tvhi/cJi  is  stqjposed  to  be  running  through  a  given 
nevei'-ending  seque?ice  of  values,  is  said  to  approach  the  number 
a  as  limit,  if  the  difference  a  —  x  tvill  tdtimately  become  and 
remain  numericallij  less  than  evenj  jjositive  number  8  tliat  we 
may  assign. 

Observe  that  it  is  not  enough  that  a  -  x  become  less  than  5  ;  it  must 
also  remain  less,  if  x  is  to  approach  a  as  limit. 

Thus,  if  X  run  through  the  sequence  i,  0,  f ,  0,  |,  0,  •  ■  • ,  the  difference 
1  -  X  will  become  less  than  every  5  that  we  can  assign,  but  it  will  not 
remain  less  than  this  5,  and  x  will  not  approach  1  as  limit. 

In  particular,  a  -  x  may  become  0;  that  is,  x  may  reach  its  limit  a. 


IRRATIOXAL   NUMBERS  59 

To  indicate  that  x  is  approaching  the  limit  a,  we  write     190 
either  a:  =  a,  read  ''ic  approaches  a  as  limit/'  or  lim  x  —  a, 
read  "  the  limit  of  x  is  a." 

Whether  a  variable  x  approaches  a  limit  or  not  depends     191 
entirely  on  the  character  of  the  sequence  of  values  through 
which  it  is  supposed  to  be  running. 

Thus,  while  x  approaches  a  limit  when  it  runs  through  the  sequence 
i»  f »  f )  •  •  •  >  plainly  it  does  not  approach  a  limit  when  it  runs  through  the 
sequence  1,  2,  3,  4,  •  •  -,  or  the  sequence  1,  2,  1,  2,  ■  ■  ■. 

Hence  the  importance  of  the  following  theorems : 

Theorem  1.     If  the  variable  x  continually  increases,  bid,  on     192 
the  other  hand,  remains  always  less  than  some  given  7iumber  c, 
it   approaches   a   limit.     And  this  limit  is   either  c    or   some 
number  which  is  less  than  c. 

For  by  hypothesis  there  are  numbers  which  x  will  never 
exceed.  Assign  all  such  numbers  to  a  class  i?2,  and  all  other 
numbers,  that  is,  all  numbers  which  x  will  ultimately  exceed, 
to  a  class  '7?i. 

We  thus  obtain  a  separation  of  the  entire  system  of  real 
numbers  into  two  parts,  B.-^,  R^,  so  related  that  each  number 
in  ill  is  less  than  every  number  in  Ro. 

Obviously  there  is  no  last  number  in  R^.  Hence,  §  160, 
there  is  a  first  number  in  R^.  Call  this  number  a.  As  x 
increases,  it  will  approach  a  as  limit. 

For  however  small  8  may  be,  if  only  positive,  a  —  8  belongs 
to  the  class  of  numbers  Ri,  which  x  will  ultimately  exceed. 
Hence  x  will  ultimately  remain  between  a  —  8  and  a,  and 
therefore  differ  from  a  by  less  than  8. 

In  the  same  manner  it  may  be  demonstrated  that 

If  the  variable  x  continually  decreases,  but,  on  the  other  hand,     193 
re7)iains  always  greater  than  some  given  number  c,  it  approaches 
a  limit.     And  this  limit  is  either  c  or  some  number  which  is 
greater  than  c. 


60  A   COLLEGE    ALGEBRA 

194  Regular  sequences.  It  is  not  necessary,  however,  that  x 
should  always  increase  or  always  decrease,  if  it  is  to  approach 
a  limit. 

Thus,  X  is  sometimes  increasing  and  sometimes  decreasing,  as  it  runs 
through  the  sequence  —  i,  |,  —  |,  1-5,  •  •  • ;  but  it  approaches  0  as  hmit. 

We  shall  prove  that  x  will  or  will  not  approach  a  limit, 
according  as  the  sequence  of  values  (Xi,  a^,  ■■•,  a„,  •••  through 
which  it  runs,  has  or  has  not  the  character  described  in  the 
following  definition : 

195  The  sequence  aj,  aj,  •  •  • ,  a„,  •  •  •  is  said  to  he  regular,  if  for 

every  positive  test  number  8  that  may  he  assigned  a  corresjiond- 
ing  term  d^  can  he  found,  which  ivill  differ  numerically  from 
every  subsequent  term  by  less  than  S. 

1.  Thus,  the  sequence  1.4,  1.4],  1.414,  •  •  •  (1),  §  161,  is  regular. 

For  the  difference  between  the  first  term,  1.4,  and  every  subsequent 
term  is  less  than  1/10;  that  between  the  second  term,  1.41,  and  every 
subsequent  term  is  less  than  1  / 10^ ;  that  between  the  nth  term  and  every 
subsequent  term  is  less  than  1/10". 

Now,  however  small  5  may  be,  we  can  give  n  a  value  which  will  make 
1/10"  smaller  still  ;  and  if  k  denote  such  a  value  of  n,  the  A;th  term  of 
1.4,  1.41,  •  •  •  will  differ  from  every  .subsequent  term  by  less  than  5. 

Thus,  if  we  assign  the  value  1/500000  to  5,  we  have  1/10^  <  5,  so  that 
the  sixih  term  of  1.4,  1.41,  •  ■  •  will  differ  from  every  subsequent  term  by 
less  than  this  value  of  S. 

2.  The  following  sequences  are  also  regular : 

h  h  h 


if,  •••, 

(2) 

h  h  h  H,  •••, 

(3) 

-h  tV,  ••• 

,         (4) 

2,  1,  1,  1,  •••. 

(5) 

Observe  that  in  (2)  each  term  is  followed  by  a  greater  term,  in  (3)  by 
a  lesser  term,  in  (4)  sometimes  by  a  greater  term,  sometimes  by  a  lesser. 

We  sometimes  encounter  regular  sequences  like  (5),  all  of  whose 
terms  after  a  certain  one  are  the  same.  Evidently  a  variable  which  runs 
through  such  a  sequence  will  ultimately  become  constant,  that  is,  will 
reach  its  limit. 

3.    The  following  sequences  are  not  regular  : 

1,2,3,4,...,  (6)  hhhh"--  (7) 


IRRATIONAL    NUMBERS  61 

For  in  (6)  the  difference  between  a  term  and  a  subsequent  one  may 
always  be'  indefinitely  great,  and  in  (7)  it  may  always  be  |,  and  therefore 
not  less  than  every  number,  1  for  instance,  that  we  can 


Formulas  for  regular  sequences.      1.    We   may    indicate   the     196 
relation  between  the  term  u,.  and  every  subsequent  term,  a  , 
by  the  formula,  §  63  : 

|a^  —  a^l  <  8  for  every  p>  k.  (1) 

2.  Again,  since  any  terms  a^  there  may  be  which  are  >  a^. 
will  lie  between  a^.  and  a^.  +  S,  and  any  which  are  <  a^  will 
lie  between  a,.  —  8  and  a^,  we  may  also  write 

Oi  —  S  <  flj,  <  %  +  8  for  every  2>  >  k.  (2) 

3.  It  follows  from  (2)  that  if  some  of  the  terms  a^  are  less, 
and  some  are  greater  than  a^.,  the  difference  between  two  of 
these  terms  may  exceed  8,  but  not  2  8. 

But  we  can  always  find  a  term,  a„  which  corresponds  to 
8/2  as  a/,  corresponds  to  8.  The  difference  between  every 
two  terms  after  a,  will  then  be  numerically  less  than  2(8/2), 
or  8;  that  is,  the  relation  between  every  two  of  these  terms 
will  be  that  indicated  by  the  formula 

\^P  ~  *-?!  <  ^  ^°^'  every  2^  >  q  >  I.  (3) 

Theorem   2.      The    variable   x   will  ajxproach   a  limit  if  the     197 
sequence  of  values  ai,  a2,  •  •  •,  a^,  •  •  •,  tliroiigh  which  it  is  supposed 
to  run,  is  a  regular  sequence. 

For  there  are  numbers  to  whose  right  x  will  ultimately 
remain  as  it  runs  through  the  sequence  aj,  a^,  ■■  ■,  a,^,  ■  ■  -.    (1) 

Thus,  if  5  and  at  have  the  meanings  above  explained,  x  will  remain  to 
the  right  of  a^.  —  5  after  it  reaches  the  value  a/,,  §  196  (2). 

Assign  all  such  numbers  to  a  class,  Ri,  and  all  other  num- 
bers • —  that  is,  all  numbers  to  whose  right  x  will  not  remain  — 
to  a  class,  R^. 


62  A    COLLEGE   ALGEBRA 

We  thus  obtain  a  separation  of  the  entire  system  of  real 
numbers  into  two  parts,  Ri,  R^,  so  related  that  each  number 
in  Ri  is  less  than  every  number  in  R^.  By  §  160,  a  definite 
number,  a,  exists  at  which  this  separation  occurs. 

Thus,  if  the  sequence  be  —  h  h  ~  h  i\'  " '  ''  ^^^  negative  rationals 
constitute  Ri,  but  0  and  the  positive  rationals,  E^ ;  and  a  itself  is  0. 

As  X  runs  through  the  sequence  (1),  it  will  approach  this 
number  a  as  limit. 

For  assign  any  positive  test  number,  8,  it  matters  not  how 
small.  Since  (1)  is  regular,  we  can  find  a  term,  «„„  §  196  (3), 
such  that 

[dp  —  a^]  <  8/2  for  every  j)  >  q  >  m.  (2) 

But  since  a  —  8/2  belongs  to  7?i,  all  the  values  of  x  after  a 
certain  one  will  lie  to  the  right  of  a  —  8/2.  And  since  a  +  8/2 
belongs  to  R^,  among  these  values  there  will  be  some  after  a„, 
which  lie  to  the  left  of  a  +  8/2  ;  for  otherwise  a  +  8/2  would 
belong  to  Ri,  since  x  would  ultimately  remain  to  its  right. 

Thus,  if  the  sequence  be  —  J,  i,  —  |,  j-\,  •  •  •,  and  5  =  jL,  all  values  of 
X  after  the/our^ft,  Jg^  Ue  between  a  —  5/2  and  a  +  5/2,  that  is,  between 
-  ^V  and  ^V- 

Let  a^  denote  such  a  value  of  x.     Then 

a-8/2<<<a  +  8/2, 

or  Ia-<|<8/2.  (3) 

From  (2)  and  (3),  since  q' >  m,  it  follows,  §§  78,  178,  that 

I  a  —  a^\  <  8  for  every  2^  >  q'. 

In  other  words,  after  x  reaches  the  value  a^  the  difference 
&  —  X  remains  numerically  less  than  8. 
Therefore  x  approaches  a  as  limit,  §  189. 

198  Conversely,  if  x  is  a}ij)roaching  a  limit,  a,  the  sequence  of 
values  ai,  a,^,  ■■■,  a^,  •••,  through  wliich  it  is  supjjosed  to  run, 
must  be  regular. 


IRRATIONAL    NUMBERS  63 

For  since  the  difference  a  —  x  will  ultimately  become  and 
remain  numerically  less  than  every  assigned  positive  number,  8, 
§  189,  we  can  choose  a^.  so  that 

1^  —  "t|  <  2/2  and  |a  —  a^^]  <  8/2  for  every  p  >  k; 

whence  [a^^  —  w^.]  <  8  for  every  j?  >  /,;. 

Hence  the  sequence  a^,  a^,  ■■■,  «„,  •  •  •  is  regular,  §  196  (1). 

We  may  combine  §§  197,  198  in  the  single  statement: 

The  sufficient  and  necessary  condition  that  a  variable  aj^proach     199 
a  limit  is  that  the  sequence  of  values  througli  which  it  is  sup- 
posed to  run  he  a  regular  sequence. 

SOME   IMPORTANT   THEOREMS    REGARDING   LIMITS 

In  the  present  section  a  and  h  will  denote  any  given  real 
numbers,  and  x  and  y  variables  which  are  supposed  to  run 
through  given  never-ending  sequences  of  values. 

The  limit  0.     From  the  definition  of  limit,  §  189,  it  imme-     200 
d  lately  follows  that 

1.  If  the  variable  x  will  ultimately  become  and  remain 
numerically  less  than  every  positive  number,  8,  that  may  be 
assigned,  then  x  approaches  0  as  limit ;  and  conversely. 

2.  If  X  approaches  a  as  limit,  then  a  —  x  approaches  0  as 
limit ;  and  conversely. 

Thus  X  approaches  the  limit  0,  as  it  runs  through  the  sequence  \^  \^\^l•••\ 
and  1  —  X  approaches  the  limit  0,  as  x  runs  through  the  sequence  |,  f ,  |,  •  •  • . 

A  variable  whose  limit  is  0  is  called  an  infinitesimal. 

Theorem   1 .     Tf  x  =  0  and  y  =  0,  and  A  and  B  remain  nwner-     201 
ically  less  than  some  fixed  number,  c,  as  x  and  y  vary,  then 
Ax  -I-  By  =  0. 

For  assign  any  positive  number,  5,  it  matters  not  how  small. 

Since  x  =  0,  x  will  ultimately  remain  numerically     <  3/2  c.      §  200,  1 


64  A   COLLEGE    ALGEBRA 

Since  y  =  0,y  will  ultimately  remain  numerically      <  5/2  c.      §  200, 1 

Hence  Ax  +  By  will  ultimately  remain  numeiically  <  2c  — ,  .■•  <  5, 

2c 
and  therefore  approaches  0  as  limit,  §  200,  1. 

Thus,  if  X  d=  0  and  y  ==  0,  then  {xy  —  o)x  +  2y  ~0. 

202  Note.  This  theorem  may  readily  be  extended  to  any  finite  number  of 
variables. 

Thus,  if  X  =  0,  2/  =  0  and  2  =  0,  then  Ax  +  By  +  Cz  =  0. 

203  Theorem  2.  The  limit  of  the  sum,  difference,  product,  quotient 
of  tivo  variables  which  approach  limits  is  the  sum,  difference, 
product,  quotient  of  these  limits :  that  is,  if  x  and  y  apjyroach 
the  limits  a,  and  h  respectively,  then 

1.    x  +  7j~a  +  b.  Z.    ory^zah. 

1.    x  —  y^a  —  h.  4.    .r/y  =  a/i,  unless  J  =  0. 

Tor,  since  a  -  x  =  0  and  6  -  ?/  =  0,  §  200,  it  follows  from  §  201  that 

^(a_x)  +  B(6-2/)  =  0.  (1) 

The  formulas  1,  2,  3,  4  may  be  derived  from  (1).     Thus, 


1. 

a  +  6  -  (X  +  y)  =  (a-  X)  +  (6  -  y)  .:  =  0, 

by{i) 

that  is. 

x  +  y  =  a  +  b. 

§  200,  2 

2. 

a  -  b  -  {X  -  y)  ^  {a  -  x)  -  {b  -  7/)  .:  =0, 

by(i) 

that  is, 

X  —  y  =  a  -  b. 

§  200,  2 

3. 

ab  -  xy  ^  {a  -  x)b  +  {b  -  y)x  .:  =  0, 

by(i) 

that  is, 

xy  =  ab. 

§  200,  2 

-\- 

X 

V  ' 

^hlhil-lh^'-^-^^-y^k- 

. -0,  by(l) 

that  is, 

x/y  =  a/b. 

§  200,  2 

204         Corollary. 

7/*  X  =  a,  then  x"  =  a". 

205         Theorem 

3.      The  limit  of  the  nth  root  of  a  var 

'able  which 

7/"  X  =  a,  then    vx  =L  Va 


IRRATIONAL   NUMBERS  65 

1.  When  a  =  0.     Assign  any  positive  number,  5. 

Since  x  =  0,  x  will  ultimately  remain  numerically  <  5".  §  200,  1 

Hence  Vx  will  ultimately  remain  numerically  <  5.  §  184 

Therefore  V^  =  0.  §  200,  1 

2.  AVhen  a  is  notO.     It  follows  from  a  later  section,  §  308,  that  x  —  a 

is  always  exactly  divisible  by  Vx  —  Va,  and  that  the  quotient  Q  does 
not  approach  the  limit  0  when  x  =  a. 

It  therefore  follows  from  §  203  (1),  by  setting  ^  =  1  /  Q  and  i?  =  0,  that 

Vx  -  Va  =::  (X  -  a)  /  Q  £=  0,  that  is,   Vx  =  Va.  §  200,  2 

RELATION  OF  THE  IRRATIONAL  NUMBERS  TO  MEASUREMENT 

Length  of  a  line  segment  incommensurable  with  the  unit.     If  a     206 

line  segment  S  be  incomviensurable  with  the  unit  segment  s,  • — • 
that  is,  if,  as  when  S  and  s  are  diagonal  and  side  of  the  same 
square,  we  can  prove  that  no  aliquot  part  of  s,  however  small, 
is  contained  in  S  exactly  —  the  definition  of  length  given  in 
§  130  does  not  apply  to  S. 

But  there  is  then  a  definite  irrational  number,  a,  which 
stands  in  the  following  relation  to  S : 

The  segments  which  are  commensurable  with  the  unit  s  fall 
into  two  distinct  classes,  those  which  are  less  than  S  and  those 
which  are  greater  than  S. 

The  rational  numbers  which  are  their  lengths,  §  130,  fall 
into  two  corresponding  classes,  which  we  may  call  A^  and  A  2. 
Every  positive  rational  belongs  either  to  A^  or  to  A^,  each 
number  in  A^  precedes  every  number  in  A^,  and,  finally,  there 
is  neither  a  last  number  in  A^  nor  a  first  in  A.^* 

There  is,  then,  §  154,  a  definite  irrational  number,  a,  which 
lies  between  all  numbers  in  A^  and  all  in  ^2-     We  call  this 

*  For  were  there  a  last  number  in  A^,  then  among  the  segments  commensur- 
able with  s  and  less  than  S  there  would  be  a  greatest,  say  S'. 

But  no  svich  sesrnient  exists.  For  according  to  the  Axiom  of  Archimedes, 
explained  in  the  following  footnote,  we  could  find  an  aliquot  part  of  s  which  is 
less  than  S  —  S' ;  and  the  sum  of  S'  and  this  part  of  s  would  be  commensurable 
with  s,  less  than  S  and  greater  than  S'. 


66  A    COLLEGE   ALGEBRA 

number  a  the  length  of  S.     We  therefore  have  the  following 
definition : 

207  The  length  of  any  segment,  S,  incommensurable  loith  the  unit, 
s,  is  that  irrational  number,  a.,  ivhich  lies  between  all  rationals 
which  are  lengths  of  segments  less  than  S  and  all  rationals 
which  are  lengths  of  segments  greater  than  S. 

Thus,  V2  is  the  length  of  the  diagonal  of  a  square  in  terms  of  the  side. 

208  If  the  length  of  S  in  terms  of  s  is  a,  we  write  S  =  as,  and 
that  whether  a  is  rational  or  irrational. 

209  Real  numbers  pictured  by  points.  As  in  the  figure  of  §  134, 
take  any  right  line  and  on  it  a  fixed  point  0  as  origin;  also 
some  convenient  unit,  s,  for  measuring  lengths.  And  by  the 
distance  from  0  of  any  point  P  of  the  line,  understand  the 
length  of  the  segment  OP  in  terms  of  s,  §§  130,  207. 

We  choose  as  the  picture  of  any  given  number,  a,  that  point,  P, 
of  the  line  whose  distance  frorn  0  is  the  nutnerical  value  of  a., 
the  point  being  taken  to  the  right  or  left  of  0,  according  as  a  is 
positive  or  negative. 

If  a  is  a  rational  number,  we  can  actually  construct  P,  §  134, 
On  the  contrary,  if  a  is  irrational,  we  usually  cannot  con- 
struct P.  We  then  assume  that  P  exists,  in  other  words,  that 
on  the  line  there  is  a  single  point,  P,  lying  between  all  points 
which  picture  rationals  less  than  a  and  all  which  picture 
rationals  greater  than  a.* 

*  This  is  not  the  place  for  a  disoussion  of  the  axioms  of  geometry ;  but 
we  may  mention  the  following  beoanse  of  their  relation  to  the  subject  of 
measurement  now  under  consideration. 

1.  Axiom  of  Archimedes.  1/  s  and  S  denote  two  line  segments  such  that 
s<S,  ice  can  alwai/s  jind  an  inter/e)',  ni,  such  that  ms>S. 

2.  Axiom  of  continuity.  Tf  all  the  points  of  a  right  line  be  separated  into 
two  classes,  R,  and  K,,  siicli  that  mrl,  j,nint  in'R^  lies  to  the  left  of  every  point 
i?iR,,  thej-e  is  either  a  last  jmiiit  in  K,  nrniirst  in  R,. 

(i)  The  Axiom  of  Avchiincdcs  is  invnlved  in  the  assumption  that  every 
line  segment  can  be  measunid.  For  the  hrst  step  in  measuring  S  in  terms  of 
s  is  to  find  an  integer,  m,  such  that  (/)X-  I)  s<S<?7is. 

(2)  The  axioms!  and  2  enable  us  to  prove  the  assumption  in  §  200  that  for 
every  given  irrational  a  there  exists  a  corresponding  point,  P. 

For  a  separates  the  rational  system  into  two  parts,  which  we  may  name  B 
and  C  respectively.    Call  the  points  corresponding  to  the  numbers  iu  each  the 


IRRATIONAL   NUMBERS  67 

Conversely,  when  P  is  given,  we  can  find  a,  at  least  approxi-    210 
mately,  by  measuring  OP  and  attaching  the  +  or  —  sign  to 
the  result,  according  as  P  is  to  the  right  or  left  of  O. 

Thus,  if  P  is  to  the  right  of  O,  and  we  can  lay  s  along  OP  five  times, 
the  tenth  part  of  s  along  the  part  left  over  seven  times,  and  the  hundredth 
part  of  s  along  the  part  still  left  over  six  times,  then  5.76  will  be  the  value 
of  a  to  the  second  place  of  decimals. 

In  this  manner  we  set  up  a  relation  of  one-to-one  correspond-     211 
ence,  §  2,  between  all  the  real  numbers  and  all  points  on  the 
line  ;  and  if  a  and  b  denote  any  two  real  numbers,  and  P  and  Q 
the  corresponding  points,  P  will  lie  to  the  left  or  right  of 
Q  according  as  a  is  less  or  greater  than  b. 

Thus,  if  a  and  b  are  positive  and  a  <  b,  and  if  c  denote  a  rational  lying 
between  a  and  b,  and  B  the  corresponding  point,  we  have,  §  206, 

OP<OR  and  OR<OQ  and  therefore  OP<OQ. 

jS-points  and  the  C'-points  respectively.     We  are  to  show  that  there  is  in  the 
line  a  definite  point,  P,  which  separates  all  the  i?-points  from  all  the  C-points. 

First  assign  the  i?-points  and. all  intermediate  points  to  a  class  i?,  and  all 
points  to  the  right  of  these  to  a  class  R,,  and  let  P  denote  the  point  which  this 
separation  defines,  by  2. 

Next  assign  the  C-points  and  all  intermediate  points  to  a  class  S^  and 
all  points  to  their  left  to  a  class  S^,  and  let  Q  denote  the  point  which  this 
separation  defines,  hy  2. 

The  points  P  a7id  Q  must  coincide.    For  if  not,  let  PQ  denote  the  line  seg- 
ment between  them.    By  1,  we  can  find  an  integer,  m,  such  that 
mPQ>s,  and  therefore  PQ>s/m. 

But  this  is  impossible.  For  we  can  select  from  B  a  number  b  and  from  0 
a  number  c  such  that  c—b<l/m.  And  if  L  and  M be  the  points  corresponding 
to  b  and  c  respectively,  we  have 

LM<s/m,  and  PQ<LM,  and  therefore  PQ<s/m. 

It  is  this  one  point,  P  or  Q,  that  corresponds  to  a  according  to  §  209. 

Q^)  Finally,  observe  that  corresponding  to  2  the  system  of  real  numbers  has 
the  property  described  in  §  KiO,  and  corresponding  to  1  the  property : 

If  a  and  b  are  any  two  positive  real  numbers,  we  ca7i  always  find  an  integer, 
m,  such  that  mb>a. 

For,  by  §§  108,  176,  178,  we  can  choose  an  integer,  m,  such  that  m>a/b  and 
therefore  ??ib>a. 

The  real  system  would  not  possess  this  property  —  at  least  not  without  a 
sacrifice  of  some  of  its  other  properties  —  were  we  to  invent  more  than  one 
irrational  for  a  separation  of  the  rational  system  of  the  kind  described  in  §  154. 

Thus,  if  every  rational  is  either  an  o,  or  an  a^,  and  a^<\)<c<a^  for  every 
Oi,  a,,  we  should  have  c-  b<a„  — a,,  §  178  and  proof  of  §  163. 

But  however  small  a  positive  number,  S,we  might  assign,  we  could  find  no 
integer,  m,  so  great  that  )/i(c  — b)>5. 

For  it  would  then  follow  that  c  — b>6/m,  which  is  impossible  since 
c-b<a2— ai  and  we  can  choose  a^,a^s,o  that  a^—a^K^/m. 


68  A   COLLEGE    ALGEBRA 

212  Theorem.      If  the  length  of  S  in  terms  of  T  is  a,  and  that  of 

T  iVi  terms  of  s  is  b,  then  the  length  of  S  in  terms  of  s  is  ab. 

1.  When  a  and  b  are  rational. 

Let  a.  —  a/h  and  b  =  c/d,  where  a,  6,  c,  d  denote  integers. 

Since  S  contains  the  6th  part  of  T  a  times,  §  130,  6S  will  contain  T 

itself  a  times,  that  is, 

6S  =  aT.  0) 

Similarly  dT  =  cs.  (2) 

But  from  (1)  and  (2)  it  readily  follows  that 

hdS  =  adT,  and  adl  =  acs, 

and  therefore  bdS  =  acs. 

That  is,  the  length  of  S  in  terms  of  s  is  —  or §  130 

bd        b    d 

2.  WLen  a  and  b,  one  or  both,  are  irrational. 

Let  Si  and  S2  denote  any  segments  commensurable  with  T,  such  that 
Si<S<S2, 
and  let  ai,  02  be  the  lengths  of  Si,  S2  in  terras  of  T,  so  that 

Si  =  aiT  and  S2  =  a2T,  where  ai<a<ff2-  §208 

Similarly,  let  Ti,  T2  denote  any  segments  commensurable  with  s,  such 
Ti<T<T2, 
and  let  61,  62  be  the  lengths  of  Ti,  T2  in  terms  of  s,  so  that 
Ti  =  bis,  and  T2  =  boS,  where  61  <b  <62- 
Then  since      Si  =  aiT,  and  T>Ti,  and  Ti  =  M, 
we  have,  by  case  1,  Si>ai6is. 

Similarly  S2  <  02628- 

Hence  ai6is<Si<S<S2  <a2?'2S, 

and  therefore  aibis  <  S  <  a^b^s. 

We  have  thus  demonstrated  that  all  the  numbers  ai^i  and  0262  are 
lengths,  in  terms  of  s,  of  segments  respectively  less  and  greater  tlian  S. 
Therefore  ab,  the  one  number  which  lies  between  all  the  numbers  ai6i 
and  02621  §  1"1»  is  the  length  of  S  itself  in  terms  of  s,  §  207. 


IRRATIONAL   NUMBERS  69 

Corollary.     If  the  lengths  of  S  and  T  in  terms  of  s  are  a  and     213 

b  respectively,  then  the  length  of  S  in  terms  ofT  is  a/b. 

For  let  the  length  of  S  in  terms  of  T  be  x. 

Then  since  the  length  of  S  in  terms  of  T  is  x,  and  that  of  T  in  terms 
jf  s  is  b,  the  length  of  S  in  terms  of  s  is  xb,  §  212. 
But,  by  hypothesis,  the  length  of  S  in  terms  of  s  is  a. 

Hence  xb  =  a, 

and  therefore  x  =  a/b. 

The  continuous  variable.     One  of  our  most  familiar  intuitions     214 
is  that  of  continuous  viotion- 


Suppose  the  point  P  to  be  moving  continuously  from  A  to  B 
along  the  line  OAB\  and  let  a,  x,  and  b  denote  the  lengths  of 
OA,  OP,  and  OB  respectively,  O  being  the  origin. 

According  to  the  assumption  of  §  209,  the  segment  AB  con- 
tains a  point  for  every  number  between  a  and  b,  through  which, 
of  course,  P  must  pass  in  its  motion  from  .1  to  B.  This  leads 
us  to  say  that  as  P  moves  continuously  from  A  to  B,  x  increases 
from  the  value  a  to  the  value  b  through  all  intermediate  values, 
or  that  X  varies  continuously  from  a  to  b. 

Of  course  it  is  impossible  actually  to  trace  the  variation  of  this  x,  since 
to  any  given  one  of  its  values  there  is  no  next  foUovring  value.  If  we 
attempt  to  reason  about  x  mathematically,  we  must  content  ourselves 
with  defining  it  thus  :  (1)  x  may  take  every  given  value  between  a  and  b, 
and  (2)  if  p  and  q  denote  any  given  pair  of  these  values,  and  p<q,  then 
X  will  take  the  value  p  before  it  takes  the  value  q.  We  may  add  that  x  is 
often  called  a  continuous  variable  when  only  the  first  of  these  properties 
is  attached  to  it. 

Ratio.     Let  M  and  N  denote  any  two  magnitudes  of  the     215 
same  kind.     By  the  vieasure  of  31  in  terms  of  N,  or  the  ratio 
of  M  to  N,  we  mean  the  very  same  numbers  which  we  have 
defined  as  lengths  in  §§  81,  130,  207,  when  M  and  N  denote 
line  segments. 


70  A   COLLEGE    ALGEBRA 

Hence  the  theorems  of  §§  212,  213  regarding  lengths  hold 
good  for  the  measures  or  ratios  of  any  magnitudes  of  the  same 
kind.     In  particular, 

216  If  the  measures  o/M  and  N  in  terms  of  the  same  unit  are  a 

and  b  respectively,  the  ratio  o/M  to  N  is  a/b. 


V.  THE  IMAGINARY  AND  COMPLEX 
NUMBERS 

PURE   IMAGINARIES 

217  The  real  system  does  not  contain  the  even  roots  of  negative 
numbers  ;  for  the  even  powers  of  all  real  numbers  are  positive. 
Thus  the  real  system  does  not  contain  the  square  root  of  —  1. 

To  meet  this  difficulty,  we  invent  a  new  system  of  signs 
called  imaginary  or  complex  numbers. 

218  The  simplest  of  these  new  signs  is  *',  called  the  unit  of 
imaginaries.  With  this  unit  and  the  real  numbers,  a,  we  form 
signs  like  ai,  which  we  then  regard  as  arranged  in  the  order 
in  which  their  "  coefficients,"  a,  occur  in  the  real  system.  We 
thus  obtain  a  new  continuous  ordinal  system  of  "numbers," 
which  we  call  pure  imaginaries. 

Thus,  proceeding  as  when  developing  the  real  system,  we  may  rirst 
form  the  complete  scale  of  imaginaries 

•  ■  •     -Si,     -  2  i,     -  i,     0,     i,     2  i,     3  /,     •  •  • , 
then  enlarge  this  into  a  de?i.se  system  by  i7itroducing  imaginaries  with 
fractional  coefficients,  and  finally  into  si  continuous  system  by  introducing 
imaginaries  with  irrational  coefficients. 

Here  2  i  is  merely  the  name  of  one  of  our  new  numbers.  Its  only 
property  is  a  definite  position  in  the  new  ordinal  system.  But  when  we 
have  defined  multiplication,  we  shall  see  that  2  i  also  represents  the  product 
2  X  J  or  t  X  2.     Similarly  every  pure  imaginary  ai. 

In  particular  we  shall  define  0  •  i  as  0.     Hence  we  write  0  for  0  i. 

Observe  that  0  is  the  only  number  which  is  common  to  the  real  system 
and  the  system  of  pure  imaginaries. 


IMAGINARY   AND   COMPLEX   NUMBERS  71 

For  these  new  numbers  we  invent  operations  which  we  call     219 
addition  and  midtijjlication.     They  are  defined  by  the  following 
equations :  - 

1.    ai  +  bi  =  (a  +  i)  i.  2.    a  ■  hi  =  bi-a  =  ahL 

3.    at  •  bi  =  —  ab. 

Thus,  3,  the  product  of  two  pure  imaginaries,  ai  and  bi,  is 
to  mean  the  real  number,  —  ab,  obtained  by  multiplying  the 
coefficients  of  ai  and  bi  together  and  changing  the  sign  of 
the  result. 

We  define  jioiver  as  in  §  136.     Thus,  (aiy  =  ai  ■  ai. 

The  system  of  pure  iynaginaries  contains  the  square  roots  of    220 
all  negative  mimbers  in  the  real  system,  namely: 

V—  1  =  i  and  v  —  a''  =  ai. 

For  i'^  =  i-i  =  \i-\i=-\.  §219,3 

Therefore,  i  is  a  square  root  of  —  1,  §  138.  We  indicate  this  root  by 
V—  1,  and  thus  have  i  =  V—  1. 

In  like  manner,  it  may  be  shown  that  —  i  is  a  square  root  of  —  1.  We 
indicate  this  root  by  —  V—  1.  

Similarly,  since  (ai)'^  =  ai-  ai  =  —  a^,  we  have  ai  —  V—  a^. 

COMPLEX   NUMBERS 

To  secure  a  number  system  which  will  contain  the  higher     221 
even  roots  of  negative  numbers,  we  invent  complex  numbers. 
These  are  expressions  like  a  +  bi,  formed  by  connecting  a  real 
number,  a,  with  a  pure  imaginary,  bi,  by  the  sign  +.     They 
are  also  often  called  imaginary  mimbers. 

Until  addition  has  been  defined  for  complex  numbers  the  expression 
a  +  6t  is  to  be  regarded  as  a  single  symbol  and  the  sign  +  as  merely  a 
part  of  this  symbol. 

Since  a  =  a  -\-  Qi  and  hi  =  0  +  hi,  real  numbers  and  pure     222 
imaginaries  are  included  among  the  complex  numbers. 


72  A   COLLEGE    ALGEBRA 

223  We  regard  the  complex  numbers  as  arranged  in  rows  and 
columns  in  such  a  manner  that  all  numbers  a  +  hi  which  have 
the  same  b  lie  in  the  same  row  and  are  arranged  in  this  row 
from  left  to  right  in  the  order  of  their  a's ;  while  all  num- 
bers which  have  the  same  a  lie  in  the  same  column  and  are 
arranged  in  this  column  from  below  upward  in  the  order  of 
their  Vs.  And  we  may  consider  any  particular  complex  num- 
ber defined  by  its  position  in  this  <' two-dimensional  ordinal 
arrangement." 

In  §  238  we  shall  explain  a  method  of  picturing  this  arrange- 
ment for  all  values  of  a  and  h.  We  may  indicate  it  as  follows 
for  integral  values  of  a  and  b. 


2  +  2i 

-l+2i 

2i 

1  +  2z 

2  +2i     ... 

2  +     i 

-  1  +     i 

i 

1+     i 

2+     i     ... 

2 

-1 

0 

2 

2-     i 

-  1  -     i 

—     i 

1  -     i 

2-     i     ... 

2-2i 

-l-2i 

-2i 

1  -2i 

2-2z     ... 

This  arrangement  may  also  be  described  as  an  ordinal  system,  §  17, 
whose  elements  are  rows  (or  columns),  each  of  which  is  itself  an  ordinal 
system  of  signs  of  the  form  a  H-  hi. 

224  Definition  of  equality.  Two  complex  num])ers  are  said  to  be 
efpial  when  they  occupy  the  same  position  in  the  two-dimen- 
sional ordinal  arrangement  just  described.     Hence, 

225  If  a.  +  bi  =  c  -f-  di,  then  a  =  c  and  b  =  d ;  and  conversely. 
In  particular,  \ia+  bi  =  0,  then  «  =  0  and  b  =  0;  and  conversely. 

Of  two  unequal  complex  numbers,  like  2  +  ?,  i  and  8  +  f,  we  cannot 
say  that  the  one  is  Ze.ss  or  greater  than  —  that  is,  ])recedes  or  follows  — 
the  other,  since  complex  numbers  do  not  form  a  sbnple  ordinal  system. 

226  Definitions  of  addition,  subtraction,  multiplication.     The  sum, 

difference,  and  lyroduct  of  two  complex  numbers  a  +  bi,  c  -f  di, 
are  to  mean  the  complex  numbers  which  form  the  second 
members  of  the  following  equations : 


IMAGINARY   AND    COMPLEX   NUMBERS  73 

1.  («  +  hi)  +  (e  +  dl)  ^{a  +  c)  +  {h  +  (T)  i. 

2.  (a  +  hi)  -{c  +  dl)  =  (a-c)  +  (b-  d)  i. 

3.  {a  +  ^'0  {c  +  di)  =  (cic  -  hd)  +  {ad  +  he)  i. 

According  to  1  and  2,  addition  and  subtraction  are  inverse  operations. 
In  particular,  by  1,  (a  +  0 1)  +  (0  +  hi)  =  (a  +  0)  +  (0  +  6)  i  =  a  +  6i ;  that 
is,  a  +  6ms  the  sum  of  a  and  6t,  according  to  the  definition  1. 

These  definitions  are  in  agreement  with  tlie  commutative,  associative, 
and  distributive  laws.  In  fact,  we  arrive  at  them  by  combining  these 
laws  with  definitions  previously  given.     Thus, 

(a  +  U)  (c  +  di)  =  {a-\-hi)c  +  {a  +  hi)  di 

=  ac  +  hi  ■  c  +  a  ■  di  +  hi  ■  di 
=  (ac  —  bd)  +.  {ad  +  be)  i,  since  i^  =  —  1. 
Corollary.     A  product  vanishes  when  a  factor  vanishes.  227 

Tor        (a  +  6i)  (f>  +  0  i)  =  (a  .  0  -  6  .  0)  +  (a  •  0  -f  6  •  0)  z  =  0. 

Division.     We  define  the  quotient  of  a.  +  hi  by  c  +  di  as  the     228 
complex  number  which  multiplied  hy  c  -\-  di  will  give  a  +  hi. 
When  c  +  di  is  not  0,  there  is  one  and  but  one  such  number, 
namely,  that  in  the  second  member  of  the  equation 

a  +  hi       ac  4-  hd       he  —  ad  . 

+ 


c  +  di       c'  +  d-'    '    c^  +  d"" 
But  when  c  +  di  is  0,  no  determinate  quotient  exists. 

For  the  product  of  the  right  member  of  this  equation  by  c  +  di   is 
a  +  6t,  as  the  reader  may  easily  verify  by  aid  of  §  226. 

We  discover  that  this  number  is  the  quotient  as  follows  : 

If  a  number  exists  which  multiplied  by  c  +  di  will  give  a  +  6i,  let  it  be 
«  + 2/i. 

Then  (x  +  yi)  (c  +  di)  =  a  +  hi.  (1) 

or  {ex  —  dy)  +  {dx  +  cij)i  =  a  +  hi.  (2) 

and  therefore  ex  —  dy  =  a  and  dx  +  aj  =  b.  (3)        §  225 

Solving  this  pair  of  equations  for  x  and  y,  we  obtain 

ac  +  bd            be  —  ad         .         .,  ,    „      „      ,,, 
X  — ,  y  — J  unless  c^  +  d^  =  0.     (4) 


74  A   COLLEGE   ALGEBRA 

Moreover,  since  (4)  is  the  only  pair  of  walues  of  x  and  y  which  will 
satisfy  (o),  the  corresponding  number  x  +  yi  is  the  only  number  which 
multiplied  by  c  +  di  will  give  a  +  hi. 

It  is  evident  from  (4)  that  when  (fl  -f  #  =  0  our  definition  of  quotient 
is  meaningless.  But  if  c-  +  d^  =  0,  both  c  =  0  and  d  —  0,  since  other- 
wise we  should  have  a  positive  number  equal  to  0.  And  if  c  —  0  and 
d  =  0,  the  divisor  c  +  diis  0. 

229  The  commutative,  associative,  and  distributive  laws.  The  oper- 
ations just  defined  evidently  include  the  corresponding  opera- 
tions with  real  numbers.  Like  the  latter,  they  conform  to  the 
commutative,  associative,  and  distributive  laivs. 

Thus,  (a  +  a'i)  (b  +  b'i)  =  ab  -  a'b'  +  (ah'  +  a'b)  i,  (1) 

and  (6  +  h'i)  {a  +  a'i)  =ba  -  b'a'  +  (b'a  +  ha')  i.  (2) 

But  the  second  members  of  (1)  and  (2)  are  equal,  by  §  177. 
Hence  (a  +  a'i)  (b  +  b'i)  =  (6  +  b'i)  (a  +  a'i). 

And  the  remaining  laws  may  be  established  similarly. 

230  Rules  of  equality.     Let  a,  b,  c  denote  any  complex  numbers. 

1.  If          a  =  b,  then  a  +  c  =  b  +  c. 

2.  If  a  +  c  =  b  -f  c,  then          a  =  b. 

3.  If          a  =  b,  then         ac  =  be. 

4.  If        ac  =  be,  then          a  =  b,  unless  c  =  0. 

1.  For  let  a  =  a  +  a'i,     b  =  ?>  +  b'i,     and  c  =  c  +  c'i. 
If  a  +  a'i  =  b  +  b'i, 

then  a  =  b  and  a'  =  h'.  §  225 

Hence  a  -{■  c  =  h  -^  c  and  a'  -\-  c'  =  h'  +  c',  §  178 

and  therefore  (a -t- c) -f- (a'  + c')  t  =  (?)  +  c)  +  (6'  + c')i,  §225 

that  is,  a  +  c  =  b  +  c.  §  226 

2.  If  a  +  c  =  b  +  c, 

we  have  a  +  c  +  (—  c)  =  b  +  c  +  (—  c),  byl 

and  therefore  '  a  =  b.  §  226 

8  and  4.  The  proofs  of  these  rules  are  similar  to  those  of  1  and  2 
respectively. 


/ 


IMAGINARY   AND   COMPLEX   NUMBERS  75 

Corollary.    If  a  product  vanish,  one  of  its  factors  must  vanish.     231 
This  follows  from  §  230,  4,  by  the  reasoning  of  §  76. 

Absolute  value  of  a  complex  number.     The  positive  real  number     232 
Vo.-  +  b'  is  called  the  absolute  or  numerical  value  of  a  -{-  bi 
and  is  represented  by  \a  +  bi\.     Hence,  by  definition, 

Thus,  \2  -\-  i\  =  V'4  +  1  =:  Vs. 

When  6  =  0,  this  definition  of  numerical  value  reduces  to  that  already 
given  for  real  numbers,  §  63.  For  a  geometrical  interpretation  of  this 
definition  see  §  239. 

We  also  say  of  two  complex  numbers  that  the  first  is  numer-     233 
ically  less  than,  equal  to,  or  greater  than  the  second,  according 
as  the  absolute  value  of  the  first  is  less  than,  equal  to,  or 
greater  than  that  of  the  second. 

Thus,  2  +  3  lis  numerically  greater  than  3  +  i. 

For  |2  +  3i|  =  VT3  and  |3  +  i|  =  VlO,  and   Vl3>VlO. 

Theorem  1 .      The  absolute  value  of  a  product  of  two  complex     234 
numbers  is  equal  to  the  product  of  their  absolute  vcClues. 
Let  the  numbers  be  a  =  a  +  a'i  and  b  =  6  +  h'i. 
"Since  ab  =  a6  -  a'b'  +  {ah'  +  a'b)  i,  §  226 


we  have  [ab|  =  V(a&  -  a'b')^  +  {ah'  +  a'h)"^.  §  232 

But  on  carrying  out  the  indicated  operations  it  will  be  found  that 

{ah  -  a'h')^  +  {ah'  +  a'6)2  =  (a^  +  a'2)  (62  +  b"^). 


Hence  ^{ab  -  a'b')^  +  {ah'  +  a'by^  =  Va2  +  a'^  ■  Vb'^  +  6'2.  §  186 

That  is,  |abl  =  la|.lbl. 

Theorem  2.      The  absolute  value  of  a  sum  of  two  complex     235 
numbers  cannot  exceed  the  sum  of  their  absolute  values. 
Employ  the  same  notation  as  in  §  234. 


Then  Va2  +  a'2  +  Vfts  +  6'2^  V(a  +^6)2  +  {a'  +  b')^      (1) 

if    a2  -f-  o'2  +  62  +  5'2  +  2  V(a2  +  a'2)  (62  +  5'2) 

>  a2  +  62  +  a'2  +  6'2  +  2  (a6  +  a'6')  §  184 


76  A  COLLEGE  ALGEBRA 


.-.  if         V'(a2  +  a'2)  (62  +  b'-^)  ^ab  +  a'b'  §  178 

.-.  if  d^b'^  +  a'26'2  +  a26'2  +  a'262  >  a262  +  a'26'2  +  2  aba'b'  §  184 

.-.  if  a26'2  +  a'262  >  2  aba'b'  §  178 

.-.if  (a6'-a'6)2>0.  (2)  §178 

But  (2)  is  always  true  since  tlie  square  of  every  real  number  is  positive 
(or  0).     Hence  (1)  is  always  true  —  which  proves  our  theorem. 

Thus,  12  +  t|  =  V5  and  |l  +  3i|  =  VlO. 

But  |(2  +  i)  +  (l  +  3i)|  =  5,  and  5<V5+VlO. 

Powers  and  roots.  1.  The  7ith.  jjoiver  of.  a  +  bi,  written 
(a  +  bi)",  is  to  mean  the  product  of  n  factors  each  ot  which 
is  a  +  bi.  It  follows  from  §  226,  3,  that  this  product  is  a 
complex  number,  as  c  +  di. 

It  may  be  proved,  as  in  §  185,  that  the  laws  of  exponents 
hold  good  for  powers  of  complex  numbers  as  thus  defined. 

2.  If  (a  +  biy  =  c  +  di,  we  call  a  -{-  bi  an  7ith.  root  of  «  +  di, 
and  we  may  indicate  it  thus,  V c  +  di. 

We  shall  prove  subsequently  that  every  given  complex 
number  c  +  di  has  n  such  wth  roots :  in  other  words,  that  in 
the  system  of  complex  numbers  there  are  n  different  numbers 
whose  7ith.  powers  equal  c  +  di. 

Thus,  since  (1/V^  +  i/^)-  =  1/2  +  2  i/2  -  1/2  =  i,  the  number 
l/V2  +  i/V2isa  square  root  of  i,  and  therefore  a  fourth  root  of  —  1. 
The  remaining  three  fourth  roots  of  —  1  are 

-l/V^  +  i/V2,     l/V2-i/V2,      -l/\/2-i/V2. 

General  conclusion.  Xo  further  enlargement  of  the  number 
system  is  necessary.  For,  as  has  just  been  pointed  out,  §§  226, 
236,  the  complex  system  meets  all  the  requirements  of  the  four 
fundamental  operations  and  evolution.  And  while  certain 
other  operations  with  numbers  have  a  place  in  mathematics, 
—  among  them  the  operation  of  finding  logarithms  of  numbers, 
§  140,  —  these  operations  admit  of  definition  by  infinite  series, 

like  Ui  +  U2  +  ?/3  -I ,  whose  terms  are  complex  numbers  ;  and 

if  such  a  series  have  a  sum,  that  sum  is  a  complex  number. 


IMAGINARY   AND   COMPLEX  NUMBERS 


77 


GRAPHICAL  REPRESENTATION  OF  COMPLEX  NUMBERS 


Complex  numbers  may  be  pictured  by  points  in  a  plane,  the     238 
points  being  called  the  graphs  of  the  corresponding  numbers. 

Take  any  two  right  lines  X'OX,  Y'OY  intersecting  at  right  angles  at 
the  origin  0  ;  also  some  fixed  unit  segment 
s  for  measuring  lengths. 

1.  We  represent  each  real  number,  a, 
by  that  point  A  on  X'OX  whose  distance 
from  0,  in  terms  of  s,  is  |a|,  §  209,  taking 
A  to  the  right  or  left  of  O,  according  as  a 
is  positive  or  negative. 

2.  We  represent  each  pure  imaginary, 
bi,  by  that  point  B  on  Y^OY  whose  dis- 
tance from  O  is  |6|,  taking  B  above  or 
below  O,  according  as  6  is  positive  or 
negative. 

3.  We  represent  each  complex  number 
a  +  bi  by  the  point  P,  which  is  obtained  by  the  following  construction  ; 

Find  A  and  B,  the  graphs  of  a  and  6i,  as  in  1  and  2.  Then  through 
A  and  B  draw  parallels  to  Y'OY  and  X'OX  respectively.  The  point  F 
in  which  these  lines  meet  is  the  graph  of  a  +  bi. 

We  call  X'OX  the  axis  of  real  numbers  and  Y'OY  the  axis  of  pun 
imaginaries. 

By  this  method  we  bring  the  system  of  complex  numbers 
into  a  relation  of  one-to-one  correspondence,  §  2,  with  the  assem- 
blage of  all  points  in  a  plane.  Moreover  we  obtain  a  complete 
representation  of  the  two-dimensional  ordinal  character  of  the 
complex  system,  §  223. 

Observe  that  the  graphs  of  all  numbers  which  have  the  same  imaginary 
part  lie  on  the  same  parallel  to  X'OX,  and  that  the  graphs  of  all  numbers 
which  have  the  same  real  part  lie  on  the  same  parallel  to  Y'OY. 

The  absolute  value  of  any  complex  number  is  the  distance  of     239 
its  graph  from  the  origin. 

For  since  the  lengths  of  OA  and  AT  in  the  figure  of  §  238  are  a  and  b 
respectively,  the  length  of  OF  is  Va^  -\-  b^  or  ja  +  6i|,  §  232. 


78 


A   COLLEGE    ALGEBRA 


240 


241 


The  graphs  of  the  sum  and  j^roduct  of  two  complex  numbers 
=  a  +  a'i,  b  =  ^  +  b'i,  may  be  found  as  follows  : 

Given  A  and  B,  the  graphs  of  a  and  b  respectively.     Join  OA  and  OB 
Y  and  complete   the   p>arallelo- 

^C  gram  OACB.      Then  C  is  the 

graph  o/  a  +  b. 

For,  draw  the  perpendic- 
ulars BB,  AE,  CF,  AG. 
Then  a,  a',  h,  b'  are  the 
lengths  of  OE,  EA,  OD,  BB 
respectively,  and  the  trian- 
gles OBB  and  AGO  are 
congruent. 


-  OE  +  EF  =  OE  +  OB  =  a  +  b,  in  length, 
-.  FG  +  GC  =  EA  +  BB  =  a'  +  b',  in  length. 
;raph  oi  a  +  b  +  (a'  +  b')  i,  or  a  +  b,  §  220,  1. 


D  E         F 

Y' 
Hence  OF  - 

and  EC  -- 

Therefore  C  is  the 

When  O,  A,  B  are  in  a  straight  line  (and  always)  C  may  be  found  by 
drawing  AC  equal  in  length  and  direction  to  OB. 

Since  OC<OA  ^  AC,  i.e.<OA+  OB,  we  have  |a  +  bl<la|  +  [bl. 

The  graph  of  the  difference  a  —  b  is  that  of  the  sum  a  +  (—  b). 

Given  A  and  B,  the  graphs  of  a  and  b,  and  let  I  denote  the  graph  of  1. 
Join  OA,  OB,  lA,  a)id  on  OB  construct  the  triangle  OBC  similar  to  OIA 
and  such  that  were  OB  turned  about  O 
until  it  lay  along  OX,  OC  ivould  lie 
along  OA.     Then  C  is  the  graph  of  ab. 

This  rule  will  be  proved  later  and 
rules  derived  from  it  for  constructing 
graphs  of  quotients  and  powers. 

When  b  =  J,  OC  is  OA  turned  90° 
"counter-clockwise"  about  0. 

It  follows  that  identical  rela- 
tions   among    complex    numbers 
may  be  translated  into  geometrical  theorems.     Hence  imagi- 
nary numbers  may  express  relatiojis  amonr/  real  things. 

Thus,  the  identity  (a  +  b)/2  =  a  -1-  (b  -  a)/2  shows  that  the  diagonals 
of  a  parallelogram  bisect  each  other;  for  the  graphs  of  (a  +  b)/2  and 
a  +  (b  -  a)/2  are  the  midpoints  of  OC  and  AB  in  the  first  figure,  §  240. 


PART   SECOND  — ALGEBRA 

I.     PRELIMINARY    CONSIDERATIONS 

ON   THE  USE  OF  LETTERS   TO   DENOTE  NUMBERS 

Constants  and  variables.     In  algebra  a  letter  is  often  used  to     242 
denote  any  nuniher  whaUoever.     Thus,  in  the  formula  ah  =  ha 
the  letters  a  and  b  denote  every  two  numbers,  the  meaning  of  the 
formula  being  :  the  product  of  any  first  number  by  any  second 
number  is  the  same  as  the  product  of  the  second  by  the  first. 

In  many  algebraic  discussions  we  find  it  convenient  to  make 
the  following  distinction  between  two  letters  which  have  this 
meaning,  as  between  the  letters  b  and  x  in  the  expression  x  -\-h. 

First.  We  regard  the  one,  b,  as  having  had  a  paHicular 
value,  but  any  that  we  please,  assigned  it  at  the  outset,  which 
it  is  then  to  retain  throughout  the  discussion.  Such  a  letter 
we  shall  call  a  known  letter  or  number,  or  a  constant. 

Second.  On  the  contrary,  throughout  the  discussion  we 
regard  the  other,  x,  as  free  to  take  every  possible  value  and 
to  change  from  any  one  value  to  any  other.  Such  a  letter  we 
shall  call  a  variable. 

Unknown  letters.     But  letters  are  also  employed  to  denote     243 
particular  numbers  whose  values   are  to   be  found.     Such  a 
letter  we  shall  call  an  miknoum  letter  or  number. 

We  are  not  at  liberty  to  assign  any  value  we  please  to  an 
unknown  letter,  as  we  are  to  a  constant  or  variable  letter. 

Thus,  in  the  equation  2  a;  —  5  =  0,  x  is  an  unknown  letter  whose  value 
we  readily  find  to  be  5/2.  In  the  expression  2x  —  5  we  may  assign  x 
any  value  we  please,  but  in  the  equation  2x  —  5  =  0  we  can  assign  x  no 
other  value  than  5/2. 

79 


80  A   COLLEGE    ALGEBRA 

244  The  choice  of  letters.  The  only  necessary  restriction  on  our 
choice  of  letters  is  that  no  single  letter  be  made  to  represent 
two  different  numbers  at  the  same  time. 

•  It  is  customary,  however,  to  represent  known  numbers  or 
constants  by  the  earlier  letters  of  the  alphabet,  as  a,  b,  c ; 
unknown  numbers  and  variables  by  the  later  letters,  as  x,  y,  z. 
Besides  simple  letters  we  sometimes  use  letters  affected 
with  accents  or  subscripts :  thus,  a',  a",  a"'  read  "  a  prime," 
"  a  second,"  "  a  third " ;  and  a^,,  a^,  a^  read  "  a  sub-null," 
"  a  sub-one,"  "  a  sub-two." 

245  On  reckoning  with  letters.  When  we  represent  numbers  by 
letters,  as  a,  b,  c,  we  can  only  indicate  the  results  of  combining 
them  by  the  operations  of  arithmetic.  Thus,  to  add  b  to  a 
merely  means  to  form  the  expression  a  +  b,  which  we  there- 
fore call  the  sum  of  a  and  b.  Similarly,  the  j^roditct  of  a  by  5 
is  the  expression  ab. 

Inasmuch  as  the  literal  expressions  thus  obtained  denote 
numbers,  we  may  reckon  with  them  by  the  operations  of 
arithmetic.  But  in  such  reckonings  we  cannot  make  any 
direct  use  of  the  values  of  the  expressions,  since  these  are 
not  given.  We  can  merely  connect  the  expressions  by  the 
appropriate  signs  of  operation  and  then  simplify  the  form  of 
the  result  by  changes  which  we  know  will  not  affect  its  value, 
no  matter  what  this  may  be. 

Now,  as  we  have  seen,  §  68,  all  the  changes  that  can  be 
made  in  the  forms  of  sums  and  products  without  affecting 
their  values  are  embodied  in  the  following  formulas : 

1.    „  -f  Z,  =  />  +  «.  2.    a  +  (/>  +  c)  =  (a  +  b)+c. 

3.    ab  =  ba.     4.    a  (be)  =  (ab)  c.      5.    a(b  +  c)=  ab  -[-  ac. 

It  may  therefore  be  said  that  the  formulas  1-5  are  practi^ 
cally  all  the  definition  of  addition  and  multiplication  that  we 
either  need  or  can  use  when  combining  literal  expressions  ;  and 
the  like  is  true  of  the  remaining  operations  of  arithmetic. 


PRELIMINARY    CONSIDERATIONS  81 

Thus,  to  add  2x  +  Sy  and  4 a;  +  5 ?/  merely  means  to  find  the  simplest 
form  to  which  the  expression  2x  +  Sy  +  {4:X  +  6y)  can  be  reduced  by 
applying  formulas  1-5,  and  adding  given  numbers.     We  thus  obtain 

2a:  +  .3?/  +  (4x  +  5?/)  =  2x  +  Sy  +  ix  +  ^y  by  2 

=  2  X  +  (3  2/  +  4  X)  +  5  ?y  =  2  X  +  (4  X  +  3  y)  +  5  ?/  by  2  and  1 
=  2x  +  4x  +  3y  +  r,y    =  (2x  +  4x)  +  (3?/ +  5^)  by  2 

=  (2  +  4) X  +  (3  +  5)  y    =  Qx  +  Sy,  the  sum  required,  by  3  and  5 

THE  FUNDAMENTAL   RULES   OF  RECKONING 

In  accordance  with  what  has  just  been  said,  we  may  regard     246 
addition,  subtraction,  multiplication,  division,  involution,  and 
evolution  as  defined  for  algebra  by  the  following  rules  and 
formulas,  which  we  shall  therefore  call  the  fundamental  rules 
of  reckoning. 

In  these  formulas — which  have  been  established  for  num- 
bers of  all  kinds  in  the  first  part  of  the  book  —  the  letters 
a,  b,  c  denote  any  finite  numbers  whatsoever,  and  the  sign  of 
equality,  =,  means  "  represents  the  same  number  as." 

Addition.     The   result  of  addinrj  6  to  a  is  the  expression     247 
a  +  b.     We  call  this  expression  the  sum  of  a  and  b.     It  has 
a  value,  and  but  one,  for  any  given  values  of  a  and  b.     In 
particular,  a  +  0  =  0  +  a  =  a. 

Addition  is  a  commutative  and  an  associative  operation  ;  that     248 
is,  it  conforms  to  the  two  laws,  §§  34,  35  : 

a  +  b  =  b  +  a,        a-\-(b  -\-c)  =  (a  +  b)+c. 

The  following  rules  of  equaUty  are  true  of  sums,  §  39  :  249 

If  a  =  b,  then  a  -\-  c  =  b  -\-  c. 

If  a  -\-  c  —  b  -\-  c,  then  a  =  b* 

Subtraction.     'Y\\\'S>  \?>\)ci%  inverse  of  addition,  %b^.     Given  any     250 
two  numbers,  a  and  b,  there  is  always  a  number,  and  but  one, 

*  I,ater  we  shall  fiud  that  this  rule  does  not  hold  good  when  c  is  infinite. 


82  A   COLLEGE    ALGEBRA 

from  which  a  can  be  obtained  by  adding  b.     We  call  this  num- 

ber  the  remainder  which  results  on  subtracting  b  from  a,  and 

we  represent  it  by  the  expression  a  —  b.     Hence,  by  definition, 

(a  —  b)  -\-  b  =  a. 

In  particular,  we  represent  0  —  b  by  —  b. 

251  Multiplication.  The  result  of  multiplying  a  by  &  is  the 
expression  ab.  We  call  ab  the  •product  of  a  by  b.  It  has  a 
value,  and  but  one,  for  any  given  values  of  a  and  b. 

In  particular,  a  •  0  =  0  •  a  =  0,  whatever  finite  value  a  may 
have. 

When  5  is  a  positive  integer,  ah  —  a  ■{-  a  ^  •■  •  tob  terms. 

252  Multiplication  is  ^.commutative  and  an  associative  operation, 
and  it  is  distributive  with  respect  to  addition ;  that  is,  it  con- 
forms to  the  three  laws,  §§  45-47 : 

ab  =  ba,       a  (be)  =  (ab)  c,       a  (b  +  c)  =  ab  -\-  ac. 

253  The  following  rules  of  equality  are  true  of  products,  §§  75,  76  : 

If    a  —  b,      then  ac  =  be. 

If  ac  =  be,    then     a  =  b,  unless  c  =  0* 

If  ac  =  0,     then     a  =  0,  or  c  =  0. 

254  Division.  This  is  the  inverse  of  multiplication,  §  124.  Given 
any  two  numbers,  a  and  b ;  except  when  b  is  0,  there  is  always 
a  number,  and  but  one,  from  which  a  can  be  obtained  by 
multiplying  by  b.  We  call  this  number  the  quotient  which 
results  on  dividing  a  by  b,  and  we  represent  it  by  the  expres- 
sion -  or  a/b.     Hence,  by  definition, 

{j\b  =  a,  except  when  i  =  0. 

255  Involution.  This  is  a  case  of  repeated  mnlflplicatio7i.  We 
represent  the  continued  product  a  ■  a  ■  ■  ■  to  n  factors  by  a"  and 
call  it  the  7ith  power  of  a. 

*  Later  we  shall  find  that  this  rule  does  uot  hold  good  when  c  is  infinite. 


PRELIMINARY    CONSIDERATIONS  83 

In  the  symbol  a",  n  is  called  the  exponent,  and  a  the  hase. 
Involution,  or  raising  to  a  power,  conforms  to  the  following     256 
thi-ee  laws,  called  the  laws  of  exponents,  §  185 : 

a™  •  a"  =  a"'  +  «,        (a'")"  =  a""*,        (ciby  =  a"b\ 

The  following  i-ules  of  equaliti/  are  true  of  powers,  §  184  :         257 

If    a  =  b,    then  a"  =  ft". 

If  a"  =  Zi-,  then    a  =  h,  or  a  =  —  b. 

The  second  of  these  rules  and  the  general  rule  of  which  it 
is  a  particular  case  will  be  demonstrated  later. 

Evolution.     This  is  one  of  the  inverses  of  involution,  §§  138,     258 
140.     Given  any  ^jostVtiJe  number  a,  there  is  a  positive  number, 
and  but  one,  whose  ?^th  power  is  equal  to  a.     We  call  this 
number  the  2)ri7ic ip a  I  nth  root  of  a,  and  we  represent  it  by  Va, 
or,  when  w  =  2,  by  Va.     Hence,  by  definition, 

(>/«)"  =  a. 

But  this  positive  number,  -Va,  is  not  the  only  number 
whose  7ith.  power  is  equal  to  a.  On  the  contrary,  as  will  be 
shown  subsequently,  there  are  n  different  numbers  whose  nth 
powers  are  equal  to  a  ;  and  this  is  true  not  only  when  a  is 
positive,  but  also  when  a  is  any  other  kind  of  number. 

When  a  is  positive  and  n  is  odd,  the  principal  nth  root  of 
—  a  is  —  va. 

On  the  reversibility  of  the  preceding  rules.     We  have  called     259 
certain  of  the  rules  just  enumerated  rules  of  equaliti/ ;  we  may 
call  the  rest  rules  of  combinatio7i. 

Observe  that  all  the  rules  of  combination  and  the  rules  of 
equality  for  sums  are  reversible,  but  that  the  rules  of  equality 
for  products  and  powers  are  7iot  completely  reversible. 

Thus,  according  to  the  distributive  law,  a{b  +  c)  =  ab  +  ac,  which  is 
one  of  the  rules  of  combination,  we  can  replace  a{b  +  c)  by  ab  +  ac,  or 
reversely,  ab  +  ac  by  a  (6  +  c). 


84  A  COLLEGE   ALGEBRA 

Again,  if  a  =  6,  we  may  always  conclude  that  a  +  c  t=b  +  c,  and 
reversely,  that,  if  a  +  c  =  6  +  c,  then  a  =  b. 

But  while,  if  a  =  6,  we  may  always  conclude  that  ac  =  bc;  on  the  con- 
trary,  if  ac  =  be,  we  can  conclude  that  a  =  b  only  when  we  know  that 
c  is  not  0. 

And  while  from  a  =  6  it  always  follows  that  a^  =  b^,  from  a^  =  62  jt 
only  follows  that  either  a  =  b  or  a  =  —  b. 

260  The  rules  of  inequality.     The  formula  a  ^  b  means  "  a  is  not 
equal  to  b." 

Of  two  given  unequal  real  numbers,  a  and  b,  the  one  is  alg/e- 
hraically  the  greater,  the  other  algebraically  the  lesser,  §  62. 
If  a  is  the  greater  and  b  the  lesser,  we  write 

a  >  b  or  b  <  a. 

In  particular,  we  have  a  >  0  or  a  <  0,  according  as  a  is 
positive  or  negative. 

261  For  any  given   real  numbers  a,  b,  c,  we  have  the  rules, 
§§178,  184: 

1.  If  a  =  b  and  b  =  c,  then  a  =  c. 
If  a  =  b  and  b  <  c,  then  a  <  c. 
If  a  <  b  and  b  <  c,  then  a  <.  c. 

2.  According  as  a  <,  =,  or  >  Z», 

so  is  a  +  c  <,  =,  or  >  ^>  -f-  r, 

and  ac  <,  =,  or   >  be,  if  c>0; 

but  ac  >,  =,  or  <.  be,  if  c  <  0. 

3.  When  a  and  b  are  j)ositive, 
according  as  a  <,  =,  or  >  J, 
so  is  a"  <,  =,  or   >  h"; 
and  Va  <,  =,  or  >  V^. 

As  has  already  been  pointed  out,  the  rules  under  2  and  3 
which  involve  only  the  sign  =  hold  good  of  imaginary  num- 
bers also.  This  is  also  true  of  the  rule  :  If  a  =  6  and  b  =  c, 
then  a  =  c,  which  we  may  call  the  general  rule  of  equality. 


PRELIMINARY   CONSIDERATIONS  85 


ADDITIONAL  ALGEBRAIC  SYMBOLS 

Besides  the  symbols  whose  meanings  have  been  explained     262 
in  the  preceding  sections,  the  following  are  often  employed 
in  algebra : 

1.  Various  signs  of  aggregatio7i,  like  the  parentheses  () 
employed  above,  and  [  ],  ^  ^,  to  indicate  that  the  expression 
included  by  them  is  to  be  used  as  a  single  symbol. 

2.  The  double  signs  ±,  read  "  plus  or  minus,"  and  q:,  read 
"  minus  or  plus." 

Thus,  in  a  ±  6  =F  c,  which  means  a  -{-h  —  c  or  a  —  h  -\-  c,  the  upper 
signs  being  read  together  and  the  lower  signs  similarly. 

3.  The  symbol  .'.  for  hence  or  therefore. 

4.  The  symbol  •  •  •  for  and  so  on. 

5.  Also,  •.•  for  since ;  >  for  not  greater  than;  <^  for  not  less 
than  ;  \  for  greater  or  less  than. 

ALGEBRAIC  EXPRESSIONS 

Any  expression  formed  by  combining  letters,  or  letters  and     263 
numbers,  by  the  operations  just  described,  is  called  an  algebraic 
expression. 

Note.     The  number  of  times  that  an  operation  is  involved  in  such     264 
an  expression   may   be   limited,  as  in   1  +  x  +  x^,  or  unlimited,  as  in 
1  +  X  +  x2  +  •  •  •,  supposed  to  be  continued  without  end.     In  the  one 
case  we  say  that  the  expression  is  finite,  in  the  other,  infinite.     For  the 
present  we  shall  have  to  do  with  finite  expressions  only. 

It  is  customary  to  classify  algebraic  expressions  as  follows,     265 
according  to  the  manner  in  which  the  variable  (or  unknown) 
letters  under  consideration  occur  in  them  : 

An  expression  is  called  integral  if  it  does  not  involve  an    266 
indicated  division  by  an  expression  in  which  a  variable  letter 
occurs  ;  fractional,  if  it  does. 


86  A   COLLEGE   ALGEBRA 


Thus, 

ifx 

and  y 

are 

the  variable  letters. 

,  but  a,  b,  c  constants, 

en 

ax'- 

-\-bx  +  c 

and 

l^v, 

are  integral. 

t 

y  + 

1 

X 

and 

2  +  x 

1  -X 

are  fractional. 

267  An  expression  is  called  rational  if  it  does  not  involve  an 
indicated  root  of  an  expression  in  which  a  variable  letter 
occurs;  irrational,  if  it  does. 

Thus,  a  +  Vbx  is  rational,  but  Vy  +  Vy  -x  is  irrational. 

268  Notes.  1.  In  applying  these  terms  to  an  expression,  we  suppose  it 
reduced  to  its  simplest  form.  Thus,  Va;2  +  2  xy  +  y'^  is  rational,  since  it 
can  be  reduced  to  the  rational  form  x  +  y. 

2.  The  terms  integral,  rational,  and  so  on,  have  nothing  to  do  with 
the  numerical  values  of  the  expressions  to  which  they  are  applied. 

Thus,  X  +  2  is  a  rational  integral  expression,  but  it  represents  an 
integer  only  when  x  itself  represents  one.  It  represents  a  fraction  for 
every  fractional  value  of  x,  and  an  irrational  number  for  every  irrational 
value  of  X. 

269  When  an  algebraic  expression  A  is  made  up  of  certain  parts 
connected  by  +  or  —  signs,  these  parts  with  the  signs  imme- 
diately preceding  them  are  called  the  terms  of  A. 

Thus,  the  terms  of  the  expression 

I  +  m 

a  +  a^c-{b  +  c)  +  [d  +  e]~  {f  +  g]  +  h  +  i+j\-  -^ 

-\-k\      ^  '^P 

are  a,  a-c,  —  (b  +  c),  and  so  on,  those  of  the  terms  which  themselves 
consist  of  more  terms  than  one  being  enclosed  by  parentheses  or  some 
equivalent  sign  of  aggregation,  §262,  1. 

270  Integral  expressions  are  called  monomials,  binomials,  trino- 
mials, and  in  general  polynomials,  according  to  the  number  of 
their  terms. 

271  In  any  monomial,  the  product  of  the  constant  factors  is 
called  the  coefficient  of  the  product  of  the  variable  factors. 

Thus,  in  4  ab'^xhj^,  4  ab'^  is  the  coefficient  of  x^y*. 

At  the  same  time,  it  is  proper  to  call  any  factor  the  coefficient  of  the 
rest  of  the  product. 


PRELIMINARY   CONSIDERATIONS  87 

In  every  monomial  the  coeflBcient  should  be  written  first.     When  no 
ooefBcient  is  expressed,  it  is  1.     Thus,  1  is  the  coefficient  of  x'^y. 

Like  terms  are  such  as  differ  in  their  coefficients  at  most.  272 

Thus,  —  2  z-y  and  hxP^y  are  like  terms. 

The  degree  of  a  monomial  is  the  sum  of  the  exponents  of  such    273 
of  the  variables  under  discussion  as  occur  in  it. 

Thus,  if  the  variables  are  x  and  y,  the  degree  of  4  ah'^x^y*  is  seven ;  that 
of  ax3,  tkree  ;  that  of  6,  zero  (see  §  595). 

The  degree  of  a  polynomial  is  the  degree  of  its  term  or  terms     274 
of  highest  degree  ;  and  the  degree  of  any  integral  expression  is 
that  of  the  simplest  polynomial  to  which  it  can  be  reduced. 

Thus,  the  degree  of  az^  +  hz-y  +  cy^  +  dx-  +  ey  +/  is  three;  and  the 
degree  of  (x  —  1)  (x  —  2)  is  two. 

It  is  convenient  to  arrange  the  terms  of  a  polynomial  in  the     275 
order  of  their  degrees,  descending  or  ascending,  and  if  there 
are  several  terms  of  the  same  degree,  to  arrange  these  in  the 
order  of  their  degrees  in  one  of  the  variables. 

This  order  is  observed  in  the  polynomial  given  in  §  274. 

A  polynomial  is  said  to  be  homogeneous  when  all  its  terms     276 
are  of  the  same  degree. 

Thus,  5  x^  —  2  x-y  +  4  xy^  +  y"^  is  homogeneous. 

Polynomials  in  a  single  variable.  Rational  integral  expres-  277 
sions  in  a  single  variable,  as  a-,  are  of  especial  importance. 
They  play  much  the  same  role  in  algebra  as  integral  numbers 
in  arithmetic.  In  fact  we  shall  find  that  they  possess  many 
properties  analogous  to  those  of  integral  numbers.  They  can 
always  be  reduced  to  the  form  of  a  polynomial  in  x,  that  is, 
one  of  the  forms  : 

UqX  -t  «!,        a^pc!^  +  a-iX  +  a.,,        a^fc^  +  a^x"^  +  a^x  +  ctg,  •  •  •, 

or,  as  we  shall  say,  to  the  form : 

ttfyX"  +  a^x"-^  -f-  a^x"-^  4 h  a„_iX  +  a„, 


88  A    COLLEGE    ALGEBRA 

where  n  denotes  the  degree  of  the  expression,  and  the  dots 
stand  for  as  many  terms  as  are  needed  to  make  the  entire 
number  of  terms  n  +  1. 

The  coefficients  Uo,  cii,  ■■■,  denote  constants,  which  may  be 
of  any  kind.  In  particular,  any  of  them  except  ao  may  be  0, 
the  polynomial  then  being  called  incomplete. 

Observe  that  in  each  term  the  sum  of  the  subscript  of  a 
and  the  exponent  of  x  is  the  degree  of  the  polynomial. 

Thus,  in  5  x^  —  x*  +  2  x2  +  X  —  3,  we  have  n  =  6,  a©  =  5,  Ui  =  0,  a-i  =  0, 
Cs  =  —  1,  04  =  2,  as  =  1,  tte  =  —  3. 

278  Functions.  Clearly  an  algebraic  expression  like  cc  +  2  or 
x^  +  y,  which  involves  one  or  more  variables,  is  itself  a  vari- 
able. We  call  X  +  2  o,  function  of  x  because  its  value  depends 
on  that  of  x  in  such  a  way  that  to  each  vahie  of  x  there 
corresponds  a  definite  value  of  x  -\-  2. 

For  a  like  reason  we  call  x^  +  y  a  function  of  x  and  y  and, 
in  general,  we  call  every  algebraic  expression  a,  function  of  all 
the  variables  which  occur  in  it. 

279  What  we  have  just  termed  integral  or  fractional,  rational 
or  irrational  expressions  in  x,  x  and  y,  and  so  on,  Ave  may  also 
term  integral  or  fractional,  rational  or  irrational  functions  of 
X,  X  and  y,  and  so  on. 

280  We  shall  often  represent  a  given  function  of  x  by  the 
symbol /(a;),  read  "function  of  x."  We  then  represent  the 
values  of  the  function  wliich  correspond  to  x  =  0,  1,  b,  hj 
/(O),  /(I),  f(h). 

Thus,  a  fix)  =  X  +  2,  we  have /(O)  =  2,  /(I)  =  3,  f(b)  =  6  +  2.  And, 
in  general,  if /(x)  represent  any  given  expression  in  x,  f(b)  represents  the 
result  of  substituting  b  for  x  in  the  expression. 

When  dealing  with  two  or  more  functions  of  x,  we  may 
represent  one  of  them  by  f(x),  the  others  by  similar  symbols, 
asF(x),<t>(x),yf;(x). 

In  like  manner,  we  may  represent  a  function  of  two  vari- 
ables, x  and  y,  by  the  symbol  f(x,  y),  and  so  on. 


PRELIMINARY    CONSIDERATIONS  89 


1.  What  is  the  degree  of  x-yz^  +  2  x^y*z^  +  3  x'^y'^z^  with  respect  to 
X,  y,  and  z  separately  ?  with  respect  to  y  and  z  jointly  ?  with  respect 
to  X,  2/,  2  jointly  ? 

2.  What  is  the  degree  of  (x  +  1)  (2  x^  +  3)  (x^  -  7)  ? 

3.  Given  3  x'^  +  x^  —  4  x*  +  x^  —  12  ;  what  are  the  values  of  n,  ao,  Oi,  •  • 
in  the  notation  of  §  277  ? 

4.  If  /(x)  =  2x3  -  x2  +  3,  find/(0),  /(-  1),  /(3),  /(8). 

5.  If  /(x)  -  {x2  -  3x  +  2)/(2x  +  5),  find/(0),/(-  2),/(6). 

6.  If  /(x)  =  X  +  V^  +  3,  find /(I),  /{4),  /{5). 

7.  If  /(x)  =  2  X  +  3,  what  is  /(x  -  2)  ?  /(x^  +  1)  ? 

8.  If  /(x,  y)  =  x3  +  X  -  ?/  +  8,  find  the  following : 

/(O,  0),      /(I,  0),      /(O,  1),      /(I,  1),     /(_2,  -3). 

IDENTICAL  EQUATIONS   OR   IDENTITIES 

If  A  denotes  the  very  same  expression  as  B,  or  one  which     281 
can  be  transformed  into  B  by  the  rules  of  reckoning,  §§  247- 
258,  we  say  that  A  is  identically  equal  to  B. 

The  notation  A  =  B  means  ".4  is  identically  equal  to  5." 

Thus,      X  (X  +  2)  +  4  is  identically  equal  to  x2  +  2  (x  +  2). 
For  X  (x  +  2)  +  4  =  (x^  +  2  x)  +  4 

=  x2  +  (2x  +  4)  =  x2  +  2(x  +  2).       §§248,252 
We  call  A  =  5  an  identical  equation,  or  identity.     Hence 

An  identical  equation  A  =  B  is  a  statement  that  a  first  expres-     282 
sion,  A,  can  be  transfornied  into  a  second  expression,  B,  by  means 
of  the  rules  of  reckoning. 

In  particular,  an  identical  equation  like  283 

3-8  +  2  =  4  +  7-14 

in  which  no  letters  occur,  is  called  a  numerical  identity. 


90  A    COLLEGE    ALGEBRA 

The  following  very  useful  theorem  is  implied  in  §  282. 

284  Theorem.  If  two  polynomials  in  x  are  identically  equal,  their 
corresponding  coefficients  are  equal ;  that  is, 

If   aox"  +  aix°-^  -\ h  a„  =  boX"  +  bjx"-'  H [-  b„, 

tlien  ao  =  bo,       Ei  =  bi,  •  •  •,  a„  =  b„. 

For  were  these  coefficients  different,  the  polynomials  would  be  different 
as  they  stand  and  the  first  could  not  be  transformed  into  the  second  by 
the  rules  of  reckoning. 

Thus,  if  ax2  +  3  X  -  3  =  2  x2  +  6x  +  0,  then  a  =  2,b  =  S,  c  =  -  S. 

If,  instead  of  being  constants,  the  coefficients  ao,  Oi,  •  •  ■ ,  bo,  6i,  •  •  • 
denote  algebraic  expressions  which  do  not  involve  x,  it  follows  from  the 
identity  aoX"  +  UiX"-^  +  ■  •  •  =  boX"  +  bix"-^  +  •  •  •  tliat  ao  =  6o,  ai  ~bi,  ■  ■  •, 
in  other  words,  that  the  expressions  denoted  by  corresponding  coefficients, 
ao  and  bo,  and  so  on,  are  identically  equal. 

285  A  similar  theorem  holds  good  of  two  identically  equal  poly- 
nomials whose  terms  are  products  of  powers  of  two  or  more 
variables  with  constant  coefficients. 

Thus,  if  a  +  6x  +  cy  +  dx2  +  exy  -\- fy-  +  ■  ■  ■ 

=  a'  +  b'x  +  c'y  +  d'x'^  +  e'xy  ■\-f'y'^  +  •  •  • , 
then  a  =  a',  b  =  b',  c  =  c',  d  =  d',  e  =  e',  f  =  /',  •  •  • . 

286  Properties  of  identical  equations.  In  algebraic  reckoning  we 
make  constant  use  of  the  following  theorems : 

Theorem  1 .  If  A  =  B,  then  B  =  A. 

For  the  process  by  which  A  may  be  transformed  into  B  is  reversible 
since  it  involves  only  rules  of  combination,  §  259.  But  the  reverse  process 
will  transform  B  into  A. 

Thus,  we  may  reverse  the  transformation  in  the  example  in  §  281. 

For  x2  +  2  (x  +  2)  =  x2  +  (2  X  +  4) 

=  (x2  +  2x)-f-4  =  x(x  +  2)  +  4.     §§248,252 
Theorem  2.    If  A  =  C  and  B  =  C,  then  A  =  B. 
For  since  B=C,  we  have  C  =  B.  by  Theorem  1 

Hence  A  =  C  and  C  =  B,  and  therefore  A  =  B. 


PRELIMINARY    CONSIDERATIONS  91 

Thus,  since  x  (x  +  2)  +  4  =  a;2  +  2  a;  +  4,  §§  248,  252 

and  x2  +  2  (X  +  2)  =  x2  +  2  X  +  4,  §§  248,  252 

we  have  .  x  (x  +  2)  +  4  =  x^  +  2  (x  +  2). 

Theorem  3.     A71  identity  remains  an  identity  when  the  same 
operation  is  ijerformed  on  both  its  members. 

This  follows  from  the  rules  of  equality,  §§  249,  253,  257. 
Thus,  a  A  =  B,  then  A  +  C  =  B  +  C,  and  so  on. 

On  proving  identities.     To  prove  of  two  given  expressions,     287 
A  and  B,  that  .1  =  B,  it  is  not  necessary  actually  to  transform 
A  into  B.     As  §  286,  2,  shows,  it  is  sufficient,  if  ive  can  reduce 
A  and  B  to  the  same  form  C. 

The  following  theorem  supplies  another  very  useful  method. 

If  from  a  supposed  identity,  A  =  B,  a  known  identity,  C  =  D,     288 
can  be  derived  by  a  reversible  process,  the  stipposed  identity 
A  =  B  is  true. 

For  since  the  process  is  reversible,  A  =  B  can  be  derived  from  C  =  D. 
Therefore,  since  C  =  D  is  true,  A  =  B  is  also  true. 

Example.     Prove  that  a  +  b  —  b  is  identically  equal  to  a. 

If  we  suppose  a  +  b  —  b  =  a  (1) 

it  will  follow  that         [{a  +  b)  -  b]  +  b  =  a  +  b.  (2)        §  249 

But  (2)  is  a  known  identity,  §  250,  and  the  step  (1)  to  (2)  is  reversible. 
Therefore  (1)  is  true. 

That  it  is  not  safe  to  draw  the  conclusion  A  =  B  unless  the  process 
from  ^  =  J5  to  C  =  D  is  reversible  may  be  illustrated  thus  : 

If  we  suppose  x  =  —  x  (1) 

it  will  follow  that  x^  =  {-  x)2.  (2) 

Here  (2)  is  true,  but  it  does  not  follow  from  this  that  (1)  is  true,  since 
the  step  (1)  to  (2)  is  not  reversible,  §  259.     And  in  fact,  (1)  is  false. 

Identity  and  equality.     It  is   important   to    remember   that     289 
identity  is  primarily  a  relation  of  form  rather  than  of  value. 
At  the  same  time. 

If  A  and  B  are  finite  expressions,  and  A  =  B,  then  A  and  B  have 
equal  values  for  all  valves  of  the  letters  which  may  occur  in  them. 


92  A   COLLEGE    ALGEBRA 

For,  by  hypothesis,  we  can  transform  A  into  B  by  a  limited  number  of 
applications  of  the  rules  a  +  b  —  b  +  a,  and  so  on.  But  a -\-  b  and  b  +  a 
have  equal  values  vi'hatever  the  values  of  a  and  b ;  and  so  on. 

The  reason  for  restricting  the  theorem  to  finite  expressions  will  appear 
later. 

Conversely,  if  A  and  B  have  equal  values  for  all  values  of  the  letters 
in  A  and  B,  then  A  =  B.     This  will  be  proved  subsequently. 

Hence  in  the  case  of  finite  expressions  we  may  always 
replace  the  sign  of  identity  of  form,  = ,  by  the  sign  of  equality 
of  value,  =,  and  when  A  =  B,  write  A  =  B.  We  shall  usually 
follow  this  practice. 

This  use  of  the  sign  =  is  to  be  carefully  distinguished  from 
that  described  in  §  325. 

ON  CONVERSE  PROPOSITIONS 

290  Consider  a  proposition  which  has  the  form 

If  A,  then  B,  (1) 

or,  more  fully  expressed :   If  a  certain  statement,  A,  is  true, 
then  a  certain  other  statement,  B,  is  also  true. 

Thus,  If  a  figure  is  a  square,  then  it  is  a  rectangle. 

If  a;  =  1,  then  x  -  1  =  0. 

291  Interchanging  the  hypothesis,  A,  and  the  conclusion,  B,  of  (1) 
we  obtain  the  converse  proposition 

If  B,  then  k*  (2) 

Thus,  the  converses  of  the  propositions  just  cited  are  : 
If  a  figure  is  a  rectangle,  then  it  is  a  square. 
If  z  -  1  =  0,  then  z  =  \. 

292  As  the  first  of  these  examples  illustrates,  the  converse  of  a 
true  proposition  may  he  false. 

*  A  proposition  like  If  A  and  B,  then  C,  which  has  a  flnvble  hypothesis,  has 
tv)0  converses:  n.imely,  If  C  aixl  B,  then  A,  and  If  A  and  C,  then  /?.  Simi- 
larly, if  there  be  a  triple  hypothesis  there  are  three  converses;  and  so  on. 


THE  fu:ndamental  operations  93 

But  the  converse  of  a  true  proposition :   li  A,  then  B,  is     293 
always  true  when  the  process  of  reasoning  by  which  the  con- 
clusion, B,  is  derived  from  the  hypothesis,  A,  is  reversible;  for 
by  reversing  the  process  we  may  derive  A  from  B,  in  other 
words,  prove  If  B,  then  A. 

The  method  of  proving  a  proposition  by  proving  its  converse 
by  a  reversible  process  is  constantly  employed  in  algebra.  An 
illustration  of  this  method  has  already  been  given  in  §  288. 

When  a  proposition  :  li  A,  then  B,  is  true,  we  call  A  a  suffi-    294 
cient  condition  of  B,  and  B  a  necessary  condition  of  A. 

Tims,  the  proposition  If  x  —  1,  then  (a;  —  1)  (a;  —  2)  =  0  is  true. 
Hence  x  =  1  is  a  sufficient  condition  tliat  (x  —  1)  (x  —  2)  =  0,  and 
(x  —  1)  (x  —  2)  =  0  is  a  necessary  condition  that  x  =  1. 

When  both  the  proposition  If  A,  then  B,  and  its  converse     295 
If  B,  then  A,  are  true,  we  say  that  A  is  the  sufficient  and  neces- 
sary  condition  of  B ;  and  vice  versa. 

Thus,  both  (1)  If  X  =  1,  then  x  -  1  =  0,  and  (2)  If  x  -  1  =  0,  then 
X  =  1 ,  are  true.  Hence  x  =  1  is  the  sufficient  and  necessary  condition 
that  X  —  1  =  0  ;  and  vice  versa. 


II.     THE    FUNDAMENTAL    OPERATIONS 

ADDITION  AND   SUBTRACTION 

Sum  and  remainder.     Let  A  and  B  denote  any  two  algebraic     296 
expressions.     By  the  sum  of  A  and  B,  and  by  the  remaiyider  to 
be  found  by  subtracting  B  from  A,  we  shall  mean  the  simplest 
forms  to  which  the  expressions   A  -\-  B  and   A  —  B  can  be 
reduced  by  aid  of  the  rules  of  reckoning,  §§  247-258. 

Some  useful  formulas.     In  making  these  reductions  the  fol-     297 
lowing  formulas  are  very  serviceable,  namely  : 

1.    a  +  1)  —  c    =  a  —  c  -\-h.        2.    a  —  (b  +  c)  =  a  ~  b  —  c. 

3.    a  +  (b  —  c)  =  a  -\-  b  —  c.        4.    a  —  (b  —  c)  =  a  —  b  -[-  c, 
5.    a  (b  —  c)  —  ab  —  ac. 


94  A   COLLEGE    ALGEBRA 

These  formulas  may  be  described  as  extensions  of  the  com- 
mutative, associative,  and  distributive  laws  to  subtraction. 
We  may  prove  1  and  2  by  aid  of  the  rule,  §  249 : 

Two  exjyressions  are  equal  if  the  resxdts  obtained  by  adding 
the  same  expressio7i  to  both  are  equal. 

1.  a-\-h  —  c  =  a  —  c-\-b. 

For  the  result  of  adding  c  to  each  member  is  a  +  Z». 
Thus,  [(a  +  6)  -  c]  +  c  =  a  +  6,  §  250 

and  (a-c)Jrb  +  c^{a-c)  +  c  +  b=:a-\-h.     §§248,250 

2.  a  —  (b  -{-  e)  =  a  —  b  —  c. 

For  the  result  of  adding  b  +  c  to  each  member  is  a. 
Thus,     [a  -  (b  +  c)]  +  (b  +  c)  =  a,  §  250 

and  a~b  —  c  +  (b  +  c)  =  a~b  —  c  +  c  +  b 

=  a-b  +  b  =  a.  §§  248,  250 

We  may  prove  3,  4,  5  as  follows : 

Since  h=(b-c)+c,  §  250 

we  have,  3.     a  +  b  —  c  =  a  +[(b  —  c)-\-  c"]—  c 

=  a+(b-c)  +  c-c  §  248 

=  a+(b-c).         by  1  and  §  250 

4.  a  —  b  +  c  =  a—  [(b  —  r)  +  f  ]  +  c 

=  a—(b  —  c)—c  +  c  by  2 

=  a-(b-c).  §250 

5.  ab  —  ac  =  a  [(i  —  c)  -\-  c']—  ac 

=  a(b-c)+  ac  -  ac  §  252 

^a(b-c).  by  1  and  §250 

Observe  that  it  follows  from  §  248  and  the  formulas  1-4 
that  a  series  of  additions  and  subtractions  may  be  performed  in 
any  order  tvhatsoeuer. 


THE    FUNDAMENTAL    OPERATIONS  95 

Thus,  a-b  +  c~d  +  e  =  a  +  c-b-d  +  e,  byl 

=  a  +  c  -  (b  +  d)  +  e  -  a  +  c  +  e  -  (b  +  d),    by  2  and  1 
=  a  +  c  +  e  —  b  —  d.  '  by  2 

Rules  of  sign.     The  "  rules  of  sign  "  which  follow  are  par-     298 
ticular  cases  of  the  formulas  3,  4,  5  just  established. 

1.    a  -\-  (—  c)  =  a  —  c.         2.       a  —  (—  c)=  a  -\-  c. 

3.        a(—  c)  =  —  ac.  4.    (—  a)  (—  c)  =  ac. 

We  obtain  1,  2,  3  at  once  by  setting  6  =  0  in  §  297,  3,  4,  5  respectively. 
We  may  prove  4  as  follows  : 

(-a)(-c)  =  (-a)(0-c)  =  (-a)0-(-a)c  §297,  5 

=  0  —  {—  ac)  =  ac.  by  2  and  3 

Rule  of  parentheses.     From  the  formulas  §  248  and  §  297,     299 
2,  3,  4,  we  obtain  the  following  important  rule  : 

Parentheses  preceded  by  the  +  sign  may  he  removed ;  paren- 
theses preceded  by  the  —  sign  may  also  be  removed,  if  the  sign 
of  every  term  within  the  parentheses  be  changed. 

Parentheses  may  be  introduced  in  accordance  with  the 
same  rule. 

Thus,  a  +  6  —  c  —  d  +  e  =  a  +  6  —  (c  +  d  —  e). 

To  simplify  an  expression  which  involves  parentheses 
within  parentheses,  apply  the  rule  to  the  several  parentheses 
successively. 

Thus,    a  -\b  -  [c  -  {d  -  e)']\  =  a  -  h  +  [c  -  {d  -  e)] 
=  a  —  b-\-c  —  {d  —  e) 
=  a  —  h  +  c  —  d-\-e. 

Of  course  the  parentheses  may  be  removed  in  any  order; 
but  by  beginning  with  the  outermost  one  (as  in  the  example) 
we  avoid  changing  any  sign  more  than  once. 


96  A   COLLEGE    ALGEBRA 

300         Rules  for  adding  and  subtracting  integral  expressions.     From 
the  formulas  of  §§  248,  252,  297  we  derive  the  rules : 

To  odd  (or  subtract)  two  like  terms,  add  {or  subtract)  their 
coefficients,  and  affix  the  common  letters  to  the  result. 

To  add  two  or  more  polynomials,  write  all  their  terms  in 
succession  with  their  signs  unchanged,  and  then  simplify  by 
combining  like  terms. 

To  subtract  one  polynomial  from  another,  change  the  sign  of 
every  term  in  the  subtrahend  and  add. 

Example  1.     Add  4  ab"^  and  —  bah- ;  also  subtract  —  5  alfi  from  4  aV^. 

We  have  4  aft'^  +  ( -  5  ah"^)  =  (4  -  5)  at^  =  -  0*2 ;  §  248 

and  4a62-(-5a62)  =  [4-{-5)]rt62  =  9a52.  §297,5 

Example  2.     Add  x^  +  ax"y  +  2  ah^  and  hx-y  —  5  ah^. 

We  have  x^  +  ax"y  +  2  a&3  +  (6x2?/  _  5  aW) 

=  x3  +  ax^y  +  2  aft'  +  bx^^y  -6ab^  §  299 

=  x^  +  ax'^y  +  bx'^y  +  2ab^-5  ab^  §  248 

=  x3  +  (a  +  5)  x-'-y  -  3  a/A  §§  252,  297,  5 

Example  3.     Subtract  2  a"b  -  ab-  +  b^  from  a^  +  a-b  +  b^. 

We  have  a'  +  a"b  +  b^  -  (2  a"b  -  ab"  +  b^) 

=  a'  +  a:-b  +  b^-2a"-b  +  ab"-  -  b^  §  299 

=  a^  -  a-b  +  ab-.  §§  252,  297 

When  the  polynomials  to  be  added  (or  subtracted)  have  like 
terms,  it  is  convenient  to  arrange  these  terms  in  columns  and 
then  to  add  (or  subtract)  by  columns. 

Example  4.  Add  a*  +  a^b  -  2  a-b"-  -  ¥  and  ab"^  +  3a262  -  a%  and 
subtract  5  a^b^  —  ab^  from  the  result. 

We  have 


a* 

+  a^?) 

-  2  a%"- 

- 

-¥ 

-a^b  +  S  aW-  + 

a65 

-  ba"-b"-  + 

a63 

0*^ 

-  4  a262  -1- 

2a65- 

^ 

THE    FUNDAMENTAL   OPERATIONS  97 

EXERCISE  n 

1.  Addiax-y,      —Qax-y,       iJbx'^y,  and  —  Sbx-y. 

2.  Add  7 a2  +  2 a  -  62,       3a  +  b'^-2a^,  and  b^-4a-ia^. 

3.  Add  3x2  _  5x  +  6,       x^  +  2x  -  8,  and  -  4x2  +  3x  -  7. 

4.  Add  4  a-3  +  a^b  -  5  6%       5  a^  -  6  a62  -  a26,       i  a^  +  10  63, 

and  6  6'  -  15  a62  -  4  a26  -  10  a^. 

5.  Subtract  4a  —  26  +  6c  from  3 a  +  6  —  c. 

6.  Subtract  2  x2  -  5  x  +  7  from  x^  +  6  x2  +  5. 

7.  What  must  be  added  to  a^  +  5  a26  to  give  a^  +  b^? 

8.  From  x^  +  y^  —  6x  +  5y  take  the  sum  of 

-2x2-Gx  +  7?/-8  and  x^  +  2x'^  -  by  +  9. 

9.  Simplify  -  (a  +  6)  +  ^  -  a  -  (2  a  -  6)  ^  -  6  (a  -  4  6). 


10.  Simplify  6 X  -  ^ 4 X  +  [2x-  (3x  +  5x  +  7  -  l)  +  3]  -8^. 

11.  Simplify  2  a  -  [4  a  -  c  +  ^  3  a  -  (4  6  -  c)  -  (6  +  3  c)  ^  -  6  c]. 

12.  Subtract  x  -  (3  ?/  +  2  z)  from  z  -  [3  x  +  (y  +  52)]. 

13.  To  what  should  x^  +  8  x  +  5  be  added  to  give  x^  —  7  ? 

14.  To  what  should  x<  -  9  x2  +  3  y  be  added  to  give  y'^  +  x  -  7  ? 

MULTIPLICATION 

Product.     By  the  product  of  two  algebraic  expressions,  A     301 
and  B,  we  shall  mean  the  simplest  form  to  which  the  expression 
AB  can  be  reduced  by  means  of  the  rules  of  reckoning. 

Of  especial  importance  in  such  reductions  are  : 

1.  The  commutative,  associative,  and  distributive  laws. 

2.  The  law  of  exponents  »"*•«"  =  a"'+". 

3.  The  rules  of  sign  : 

a(—  b)  =  (—  a)b  =—  ab;       (—  a)  (—  b)  =  ab. 

Rules  for  multiplying  integral  expressions.     1.    To  find  the     303 
product  of  two  monomials,  multiply  tJte  product  of  the  numerical 


98  A    COLLEGE    ALGEBRA 

factors  hy  that  of  the  literal  factors,  simplifying  the  latter  by 
adding  exponents  of  powers  of  the  same  letter. 

Give  the  result  the  +  or  —  sign,  according  as  the  monomials 
have  like  or  unlike  signs. 

2.  To  find  the  product  of  a  polynomial  hy  a  moyiomial  or 
polynomial,  mnltiply  each  term  of  the  multiplicand  by  each 
term  of  the  multiplier  and  add  the  products  thus  obtained. 

The  first  rule  follows  from  the  commutative  and  associative  laws  and 
the  law  of  exponents.  The  second  rule  follows  from  the  distributive 
law ;  thus, 

(a  +  6  +  c)  (??i  +  n)  =  (a  +  b  +  c)  ??i  +  (a  +  6  +  c)  n 

=  am  +  hm  +  cm  +  an  +  6n  +  en. 

The  first  rule  applies  also  to  products  of  more  than  two 
monomials.  When  an  odd  jiiimber  of  these  monomials  have 
—  signs,  the  sign  of  the  product  is  — ;  otherwise  it  is  +• 

A  product  of  7}iore  than  two  polynomials  may  be  found  by 
repeated  applications  of  the  second  rule. 

Example  1.     Find  the  product  of  -  4  a"h-x^,       2  hx'^,  and  -  3  a^x. 
We  have  -  4  a^b'^x^  •  2  6x*  •  -  3  a^x  =  24  aWx'-hx^a'^x  =  24  a^b^xs. 

Example  2.     Find  the  product  of  a  —  2  6  and  ab  —  b^  +  a^. 
For  convenience  we  arrange  both  factors  in  descending  powers  of  a, 
and  choose  the  simpler  factor  as  multiplier.     We  then  have 

(a2  +  ab-  b-)  (a  -  2  6)  =  a^  +  a"-b  -  ab"-  -2a%-2  ab"-  +  2  63 
=  a3  -  a'^b  -  3  ab"^  +  2  b^. 

303  The  degree  of  the  product  Avith  respect  to  any  letter  (or  set 
of  letters)  is  the  sum  of  the  degrees  of  the  factors  with  respect 
to  that  letter  (or  set  of  letters). 

This  follows  from  §302,  1,  and  the  fact  that  the  term  of  highest 
degree  in  any  product  is  the  product  of  the  terms  of  highest  degree  in 
the  factors. 

Thus,  the  degrees  of  x2  +  1  and  x^  -  1  are  two  and  three  respectively, 
and  the  degree  of  the  product  (x^  +  1)  (x^  -  1),  or  x^^  +  x^  -  x^  -  1, 
is  five. 


THE, FUNDAMENTAL   OPERATIONS  99 

When  both  factors  are  homogeneous,  §  276,  the  product  is     304 
homogeneous. 

For  if  all  the  terms  of  each  factor  are  of  the  same  degree,  all  the 
products  obtained  by  multiplying  a  term  of  the  one  by  a  term  of  the 
other  are  of  the  same  degree.  Hence  the  sum  of  these  products  is  a 
homogeneous  polynomial. 

Arrangement  of  the  reckoning.     When  both  factors  are  poly-    305 
nomials   in  x  or  any   other   single  letter,  or  when  both  are 
homogeneous    functions   of  two   letters,   it  is    convenient   to 
arrange  the  reckoning  as  in  the  following  examples. 

Example  1.     Multiply  2x3  -  a;2  +  5  by  a;  -  3  +  x^. 
2   3  _      2  4.  i^  ^^  arrange  both  factors  in  descend- 

2   ,      a-   _  Q  ^°S  (or  ascending)  powers  of  x  and 

2^5  _ — --^ '- rJi  place  multiplier  under  multiplicand. 

2   4  _      3      '  c  ^6  tli^n  write  in  separate  rows  the 

_„3„2  _ir;     "partial  products"  corresponding  to 

o    .   ,  — I _    ,   ,   Q    >,  ,   , Tz     the  several  terms  of  the  multiplier, 

2  x^  +     x*  -  7  x3  +  8  x2  +  5  X  —  15       ,     .        ,  ,       ,  ,  ^       ' 

placing  them  so  that  like  terms,  that 

is,  terms  of  the  same  degree,  are  in  the  same  column. 

Finally  we  add  these  like  terms  by  columns. 

Example  2.     Multiply  x-  —  ?/2  +  2  xy  by  2  y  +  x. 
x2  +  2x?/  —y"  In  this  case  both  factors  are  homogeneous 

X  4-  2?/ functions  of  x  and  y. 

x3  +  2  Oo^y  —     xy'^  We  arrange  them  both  in  descending  powers 

2  x^y  +  4  xy'^  —  2  y^     of  x,  and  therefore  in  ascending  powers  of  y 
x^  +  4x-2/  +  3x2/^  —  2  y3     and  then  proceed  as  in  Ex.  1. 

Detached  coefficients.  In  the  reckoning  illustrated  in  §  305,  306 
Ex.  1,  the  terms  are  so  arranged  that  their  positions  suffice 
to  indicate  Avhat  powers  of  x  occur  in  them.  We  may  make 
use  of  this  fact  to  abridge  the  reckoning  by  suppressing  x 
and  writing  the  coefficients  only,  and  it  is  always  worth 
while  to  do  this  when  the  given  polynomials  have  numerical 
coefficients. 

If  either  polynomial  is  incomplete,  care  must  he  taken  to 
indicate  every  niissinrj  term  hij  a  ()  coefficient. 


100  A  COLLEGE  ALGEBRA  . 

Example.     Multiply  x^  -  3 12  +  2  by  x^  +  3  a;2  -  2. 

We  arrange  the  reckoning  as  in  §  305, 
1  —  d  +  0  +  j,^  .    ^^^  write  the  coefficients  only,  indi- 

X  -f  o  -r  u _  eating  the  missing  terms  by  0  coefficients. 

1  —  d  +  0  +  2  ^g  ^^j^  jj^g  partial  product  correspond- 

3  -  9  +  0  +    6  ing  to  the  0  term  of  the  multiplier.     Insert- 

— — — — — — —     ing  the  appropriate  powers  of  x  in  the  final 

1  +  0  —  9  +  U+  —  —  result  —  beginning  with  x^  since  the  sum 
of  the  degrees  of  the  factors  is  six  — we  obtain  the  product  required, 
namely,  x6-9x-*  +  12x2-4. 

The  degree  of  the  product,  .six,  is  also  indicated  by  the  number  of 
terms,  seven,  in  the  result  1+0-9  +  0  +  12-0-4,  §  277. 

This  is  called  the  method  of  detached  coefficients.  It  applies 
not  only  to  polynomials  in  a  single  letter,  —  both  arranged  in 
descending  or  ascending  powers  of  that  letter,  —  but  also  to 
homofjeneous  polynomials  in  two  letters.  For  in  arranging  two 
such  polynomials  in  descending  powers  of  one  of  the  letters, 
we  at  the  same  time  arrange  them  in  ascending  powers  of  the 
other  letter,  so  that  the  position  of  any  coefficient  will  indicate 
what  powers  of  both  letters  go  with  it. 

307         Formulas  derived  by  the  method  of  detached  coefficients.     Con- 
sider the  following  examples. 

Example  1.     Prove  the  truth  of  the  identity 

(a*  +  a^b  +  am  +  ah^  +  6*)  (a  -  6)  =  a^  -  fts. 

We   perform  the  multiplication  indicated  in 
the  first  member  by  detached  coefficients,  and 

r  so  obtain  the  coefficients  of  the  product  arranged 

1     ^"   descending   powers   of   a  and   in   ascending 

powers  of  h. 

1+0  +  0  +  0  +  0-1  ^^  ^^^^  .^  advance  that  the  degree  of  the 

product  is  five,  which  is  also  indicated  by  the  number  of  terms,  six,  in 
the  final  result.     Hence  the  product  is 

a^  +  0  •  a*6  +  0  •  aW  +  0  •  a26»  +  0  •  a6*  -  U",  or  a^  -  U'. 
Example  2.     Prove  the  truth  of  the  identities 

(a2  -  o6  +  &2)  (a  +  6)  =  a8  +  68.  (1) 

(cfi  -  cC^h  +  a62  -  63)  ^a^-h)  =  a*  -  6*.  (2) 


THE    FUNDAMENTAL   OPEIlATlCNS  101 

Proceeding  precisely  as  in  Ex.  1,  we  have 

1  -  1  +  1     (1)  1  -  1  +  1  -  1     (2) 

1 +1  1 + 1 

1-1+1  1-1+1-1 

1-1+1  1-1+1-1 

1  +  0  +  0  +  1,  i.e.  0=5  +  63.  1+0  +  0  +  0-1,  i.e.  a'  -  ¥. 

By  the  method  illustrated  in  these  examples  we  may  prove 
the  truth  of  the  following  identities,  of  which  the  examples 
are  special  cases,  namely  : 

For  eve7'i/  positive  integral  value  of  7i  we  have  308 

(a"-i  +  a"-26  -\ h  ab"-^  +  ft"  "i)  (a  -  ft)  =  a"  —  ft", 

For  every  positive  odd  value  of  7i,  we  have  309 

(a«-i  -  a'-zft  -\ aft"-2  +  ft"-')  (a  +  ft)  =  a"  +  ft". 

And  for  every  positive  even  value  of  7i,  we  have  310 

(a«-i  -  a^-H  4 h  aft"~^  -  ft""')  (a  +  b)  =  a"  -  ft". 

Powers  of  a  binomial.     We  can  compute  successive  powers     311 
of  a  +  ft  by  repeated  multiplications.     These  multiplications 
are  readily  performed  by  detached  coefficients. 

As  the  coefficients  of  the  multiplier  are  always  1  +  1,  it  is 
only  necessary  to  indicate  for  each  multiplication  the  partial 
products  and  their  sum.     We  thus  obtain 


(1)  1  +  1 

i.e.  a  +  b. 

1  +  1 

(2)  1  +  2  +  1 

i.e.  a2  +  2a6  +  62 

=  (a  +  6)2. 

1+2  +  1 

(3)  1  +  3  +  3  +  1 

i.e.  a'  +  3  a26  +  3  a62  +  63 

=  (a  +  6)3. 

1+3+3+1 

(4)  1  +  4  +  6  +  4  +  1      i.e.  a*  +  4  d^b  +  6  a262  +  4  06'  +  6*       =  (a  +  6)*. 

Observe  that  in  each  multiplication  the  coefficients  of  the 
second  partial  product  are  those  of  the  first  shifted  one  place 
to  the  right.     Hence  when  we  add  the  coefficients  of  the  two 


102  A   COLLEGE   ALGEBRA 


pai'tiAl  'products  and  so'  obtain  the  coefficients  of  the  next 
power,  we  are  merely  applying  the  rule  : 

To  any  coefficient  in  a  power  already  found  add  the  coefficient 
which  precedes  it ;  the  sum  will  be  the  correspondiny  coefficient 
in  the  next  power. 

All  the  coefficients  of  this  next  poiver,  except  the  first  and  last, 
can  be  found  by  this  rule  ;  these  are  1  and  1 

Thus,  the  coefBcients  of  (4)  which  correspond  to  3,  3,  1  in  (3)  are 
3  +  1  or  4,  3  +  3  or  6,  1  +  3  or  4. 

Applying  the  rule  to  (4),  we  obtain  4  +  1  or  5,  6  +  4  or  10,  4  t-  6  or 
10,  1  +  4  or  5.     Hence 

(a  +  6)5  =  a5  +  5  a*6  +  10  a%^  +  10  a%^  +  5  a¥  +  h^. 

Evidently  the  coefficients  of  any  given  power  of  a  +  ^  can 
be  obtained  by  repeated  applications  of  this  rule. 

Example.     Find  successively  (a  +  6)^,  (a  +  by,  (a  +  h)^. 

Products   of   two   binomial   factors  of  the   first  degree.     The 

student  should  accustom  himself  to  obtaining  products  of  this 
kind  by  insjiection.     We  have 

{x  -{-  a)  {x  +  b)  =  X-  +  {a  +  b)x  -\-  ab.  (1) 

(aoX  +  a{)  (boX  +  b{)  =--  a^boX^  +  (ao^i  +  Oi^o)  ^  +  a  A.      (2) 

In  the  product  (1)  the  coefficient  of  x  is  the  snm  and  the 
final  term  is  the  product  of  a  and  b. 

In  the  product  (2)  the  first  and  last  coefficients  are  products 
of  the  first  coefficients  and  of  the  last  coefficients  of  the  factors, 
and  the  middle  coefficient  is  the  sum  of  the  ''  cross-products  " 
a^x  and  tti^o- 

Example  1.     Find  the  product  (x  +  5)  (x  —  8). 

(X  +  5)  (X  -  8)  =  x2  4  (5  -  8)  X  -  40  =  x2  -  3  X  -  40, 

Example  2.     Find  the  product  (x  +  3y)  (x  +  10?/). 

(X  +  3?/)  (X  +  lOy)  =  x2  +  (3  +  10)X7/  +  SOy^  =  x2  +  \Zxy  +  Z^y\ 


THE    FUNDAMENTAL   OPERATIONS  103 

Example  3.     Find  the  product  (2  x  +  3)  (4x  +  7). 
(2x  +  3)(4x  +  7)  =  2-4x2  +  (2.7  +  3-4)x  +  3-7  =  8x2  +  26x  +  21. 

Example  4.     By  the  methods  just  explained  find  the  products 
(X  -  10)  (X  -  15),  (3a +  46)  (5  a -66),  {7 x  -  y)(5x  -  3y). 

Product  of  any  two  polynomials  in  x.     Consider  the  product     314 

(aoX^  +  aix'^  +  a„x  +  a^  {1)qX-  +  h-^x  +  h<^ 

+  (Cfi^>2  +   ajji  +  aj)^)  X'^  +  («2^''2  +  «3^l)  ^  +  «3^2- 

The  product  is  a  polynomial  in  x  whose  degree  is  the  sum 
of  the  degrees  of  the  factors.  And  the  coefficient  of  each  term 
may  be  obtained  by  the  following  rule,  in  which  a^  denotes 
one  of  the  numbers  a^,  a^,  a^,  a^,  and  bf.  one  of  the  numbers 
^0)  ^1)  ^2-  Find  the  difference  between  the  degree  of  the  product 
and  the  degree  of  the  term,  and  then  form  and  add  all  the 
products  aj^b^.  in  which  h  +  k  equals  this  difference. 

Thus,  to  obtain  the  coefficient  of  x-,  we  find  the  difference  5  —  2,  or  3, 
and  then  form  and  add  0162,  0261,  0360,  these  being  all  the  products 
ahbk  in  which  h  +  k  =  Z. 

This  rule  applies  to  the  product  of  any  two  polynomials  in 
X  of  the  form  a^x"'  +  •  ■  •  +  «,„  and  b^yX'^  -\ +  ^„.  It  also  indi- 
cates how  to  obtain  any  particular  coefficient  of  the  product 
when  the  factors  have  numerical  coefficients. 

Example  1.     Find  the  coefficient  of  x^""  in  the  product 
{aoa:'5  +  a^x'^^  +  ■■■  +  a^x  +  075)  (60x0  +  6ix59  +  •  •  •  +  659X  +  beo). 

The  degree  of  the  product  is  75  +  60  or  135  ;  and  135  -  100  =  35. 
Hence  the  coefficient  of  x^'"'  is  00635  +  01634  +  •  •  •  +  03461  +  03560. 

Similarly  the  coefficient  of  x^  is  O40660  +  O41659  +  ■  •  •  +  O74626  +  075635. 

Example  2.     Find  the  coefficient  of  x^  in  the  product 

(3  X*  -  2  x3  +  x2  -  8  X  +  7)  (2  x3  +  5  X-  +  6  X  -  3). 
The  required  coefficient  is  (-  2)  (-  3)  +  1  •  6  +  (-  8)  5  +  7  •  2,  or  -  14. 


104  A   COLLEGE    ALGEBRA 

Example  3.  In  the  product  of  Ex.  1,  find  the  coefficients  of  x"* 
and  of  x23_ 

Example  4.  In  the  product  of  Ex.  2,  find  separately  the  coefficients 
of  x^,  x^,  X*,  x2,  and  x. 

Products  found  by  aid  of  known  identities.  The  following 
formulas  or  identities  are  very  important  and  should  be  care- 
fully memorized. 

(a  +  by  =  a''  +  2ab  +  b\  (1) 

(a  -hf  =  0^-2  ah  +  h'.  (2) 

(a  +  h)  {o  -b)  =  a'-  h\  (3) 

To  this  list  may  be  added  the  formulas  given  in  §§  308,  309, 
310,  and  the  following,  §  311 : 

'  {a^hf  =  a?^?>  a^b  ■{- ^  ab'' +  b^.  (4) 

Inasmuch  as  the  letters  a  and  b  may  be  replaced  by  any 
algebraic  expressions  whatsoever,  these  formulas  supply  the 
simplest  means  of  obtaining  a  great  variety  of  products.  The 
following  examples  will  make  this  clear. 

Example  1.     Find  the  product  (.3x  —  5?/)2. 
(3x  -  52/)2  =  (3x)-i  -  2  ■  3x  •  5?/  +  (5  (/)2  =  9x2  -  30x2/  +  25t/2.        by  (2) 

Example  2.     Find  the  product  {x2  +  xy  +  y-)  {x2  -  xy  +  y"). 
(x2  +  x?/  +  2/2)  (x2  -xy  +  2/2)  =  [(a;2  +  ^2)  +  ^y^  |-(a;2  +  yi)  _  ^y] 

=  (x2  +  2/2)2  _  y.2yi  =  a;4  +  x^^  +  y*.  by  (3),  (1) 
Example  3.     Explain  the  step.s  in  the  following  process. 
(X  +  2/  +  2)  (X  -  2/  +  2)  (X  +  2/  -  2)  (X  -  2/  -  2) 

=  [x  +  (2/  +  z)-]  [X  -  (2/  +  2)]  .  [X  +  (2/  -  2)]  [x  -  (2/  -  2)J 

=  [X2  -  (2/  +  2)2]  .  [X2  -  (2/  -  2)2] 

=  [(X2  -  2/2  -  22)  -2yZ-\-  [(X2  _  2/2  _  22)  +  2  2/2] 

=  [X2  -  (2/2  +  22)]2  _  4  y2z2 

=  X*  -  2  X2  (2/2  +  22)  +  (2/2  +  22)2  _  4  ^2^2 

=  X*  +  2/*  +  2*  -  2  x22/2  -  2  2/222  _  2  22x2, 


THE    FUNDAMENTAL    OPERATIONS  105 

Observe  in  particular  that  by  this  method  we  may  derive 
from  (1)  and  (4)  the  square  and  cube  of  any  polynomial. 

Thus,  we  have 
(a  +  6  +  c)2  =  [(a  +  6)  +  c]2  =  (o  +  6)^  +  2  (a  +  6)  c  +  c2 

=  a-  +  62  ^  c2  +  2  a6  +  2  ac  +  2  6c. 
(a  +  5  +  c)3  =  (a  +  6)3  +  3  (a  +  b)-c  +  3  (a  +  6)  c2  +  c^ 

=  a3  +  63  +  c3  +  3a'i6  +  362a  +  362c  +  3c26  +  3c2a  +  3o2c  +  6a6c. 

Generalizing  the  first  of  these  results  we  have  the  theorem  : 

The  square  of  any  polynomial  is  equal  to  the  sum  of  the     316 
squares  of  all  its  terms  together  with  twice  the  products  of  every 
two  of  its  terms. 

Example  1.     Find  the  product  (a-6  +  2c-3  d)2. 
Example  2.     Find  the  product  (1  +  2x  +  3x2)2. 
Example  3.     Find  the  product  (x^  —  o:;hj  +  xy-  —  y^)^. 

Powers  of  monomial  products.     By  the  ?2th  power  of  any     317 
algebraic   expression,  A,  we    shall   mean    the   simplest  form 
to  which  the  expression  .4"  can  be  reduced  by  the  rules  of 
reckoning. 

From  the  laws  of  exponents  (a'")"  =  a'""  and  (ab)"  =  a"!/"  we 
derive  the  following  rule  : 

To  raise  a  monomial  expression  A  to  the  nth  power,  raise  its  5(^18 
numerical  coefficient  to  the  nth  power  a?id  multiply  the  exponent  ^~~ 
of  each  literal  factor  by  n. 

If  the  sign  of  A  be  — ,  gii'e  the  result  the  sign  +  or  —,  accord- 
ing as  n  is  even  or  odd. 

Thus,  { -  2  ax2(/7)4  =  ( -  2Ya*z«y^-»  =  16  a*x«y^«. 

For  by  repeated  applications  of  the  law  {ab)"  =  a"6«  we  have 
(-  2  ax2?/7)4  =,  (_  2)ia*  (x^)*  {y-)\ 
and  by  repeated  applications  of  the  law  {a'")"  =  a™"  we  have 
(-  2)4a4(x2)4  (2/7)4  =  l6a*x82/2^. 


106  A   COLLEGE   ALGEBRA 

EXERCISE  m 

In  the  following  examples  perform  each  multiplication  by  the  most 
expeditious  method  possible.  In  particular  employ  detached  coefficients 
where  this  can  be  done  with  advantage  ;  also  the  identities  of  §  315. 

1.  Multiply  3x5  -  2x1  -  x3  +  7x2  -  6x  +  5  by  2a;2  -  3x  +  1. 

2.  Multiply  5  x3  -  3  ax2  +  2  a-'x  +  a^  by  3  x2  -  ax  -  2  a^. 

3.  Multiply  x5  -  x^t/  +  x^y-  -  x-y^  +  xy*  -  y^  hy  x  +  y. 

4.  Multiply  3  x3  -  2  x2  +  7  by  2  x3  -  3  X  +  5. 

5.  Multiply  7  X  —  2  2/  by  4  X  —  5  ?/,  by  inspection. 

6.  Multiply  a2  -  ax  +  6x  —  x2  by  6  +  x. 

7.  Multiply  X*  —  2  X  +  5  x2  -  x^  by  3  +  x2  —  x. 

8.  Multiply  2  x"  -  3x«-2  +  5x»-3  by  x»-2  -  x^-s. 

9.  Multiply  a2  _  a&  +  3  62  by  a2  +  a6  -  3  62. 

10.  Multiply  x  +  32/-2zbyx-3y  +  2z. 

11.  Multiply  x2  +  xy  +  2/2  +  X  —  y  +  1  by  X  —  ?/  —  1. 

12.  Multiply  a2  +  62  +  c2  +  6c  +  ca  -  a6  by  a  +  6  -  c. 

13.  Multiply  3x-22/  +  5byx-4y  +  6. 

14.  Multiply  x  +  7?/-32:by2x  +  ?/-8  2. 

15.  Find  the  product  (6  +  x)  (6  -  x)  (62  +  x2). 

16.  Also  (x2  +  X  +  1)  {x2  -  X  +  1)  (X*  -  x2  +  1). 

17.  Also  {x  +  y  +  z)  {-  X  +  y  +  z)  {x  -  y  +  z)  {x  +  y  —  z). 

18.  Form  a  table  of  the  coefficients  of  the  first  four  powers  of  x2  +  x  +  1, 

19.  Continue  the  table  of  coefficients  of  successive  powers  of  a  +  6 
as  far  as  the  tenth  power. 

20.  Find  (4  x  -  3  y)-  and  (4  x  -  3  y)^ 

21.  Find  (x  +  2?/  +  3z-4 m)2. 

22.  Find  (x  +  2y  +  Sz)^;  also  (x  +  2y  -3 2)8. 

23.  Multiply  (a  +  2  6)2  by  (a  -  2  6)2. 

24.  Find  the  coefficients  of  x29  and  of  x^^  in  the  product 

(aoX2T  +  aiX26  +  .  ■  •  +  02635  +  Uii)  (6oxi9  +  biX^»  +  ••'  +  bi»x  +  619). 


THE    FUNDAMENTAL   OPERATIONS  107 

25.  Find  the  coefficients  of  x^,  x^,  and  x*  in  the  product 

(2x6  -  3x5  +  4x*  -  7x3  +  2x  -  5)  ^g^^  _  x3  +  2x2  +  3x  -  8). 

26.  Verify  the  following  identities  : 

1.  (x  +  y  +  2)3  -  (x3  +  ys^z3)  =  3iy  +  z) {z  +  x) (x  +  y). 

2.  {a^  +  bf^)  (x2  +  2/2)  =  (ax  +  byf  +  (6x  -  ay)\ 

3.  (a2  -  62)  (x2  _  yi)  =  (ax  +  by)^  -  (bx  +  ay^. 

4.  (a  +  6  +  c)3  =  a3  +  63  +  c3  +  3a2{6  +  c)  +  362(c  +  a) 

+  3c2(a  +  6)  +  6a6c. 
VH.    Simplify  the  following  powers : 

(2  a2x32/'')5,      (-x^y^zy,      (a26"'c3)2»,      (a^ftncS")". 
28.    Simplify  the  following  products  : 

( -  a62c3)  (a36)2  ( _  acZ)h^      ( _  2  x'^y^f  (axSyiija. 

DIVISION 

Quotient.     Let  A  and  B  denote  any  two  algebraic  expressions     319 
of  wliich  B  is  not  equal  to  0.     By  the  quotient  of  A  divided  by 
B^  we  shall  mean  the  simplest  form  to  which  the  fraction  A  j B 
can  be  reduced  by  the  rules  of  reckoning. 

Formulas.     In  making  such  reductions  the  following  formu-     320 
las  are  especially  useful,  namely, 

1  ^  _  ^ 

n        *'"  1  a"*  1  , 

z.      —  =  a™"",  when  m>  n\  —  = >  when  n>  m. 

a"  a"       a"~"' 


3. 
4. 

—  a           a            a             a 

b  ~~ V      -h~~V 

a  +  b       a       b 
d      ~  d^  d 

—  a       a 

We  may  prove  1,  3,  and  4  by  aid  of  the  rule,  §  253 : 

Tivo  expressions  are  equal  if  their  products  by   any  third 
expression  (not  0)  are  equal. 


108  A    COLLEGE   ALGEBRA 

For  in  1  the  product  of  each  member  by  be  is  ac. 

Thus,  —  6c  =  ac  :   and  -  be  =  ~  b  ■  c  -  ac.  §§  254,  252 

be  b  b 

Again,  in  3  the  product  of  each  member  of  the  first  equa- 
tion by  h  is  —a,  and  the  products  of  each  member  of  the 
second  and  third  equations  hj  —  b  are  a  and  —  a  respectively. 

Thus,  -^6  =  -a;and       {--\b  =  -~b=-a.     §§298,254 

6  \     b  /  b 

Finally,  in  4  the  product  of  each  member  hj  d  i?,  a  -\-  b. 

Thus,  "^^d  =  a  +  b;  f"  + -V^  =  "d  +  ^  cZ  =  a  +  6.        §§254,  252 
d  \d      d/         d        d 

The  formula  2  is  a  particular  case  of  the  formula  1. 

Thus,  if  m>  n,  a™  =  a"'-"  ■  a".  §  256 

am     a"'—"  -a"  ,     , 

Hence  —  = =  a™-".  by  1 

a"  a" 

321  Rules  for  simplifying  A/B.      The  formulas  1,  2,  and  3  give 
us  the  following  rules  for  simplifying  A/B. 

1.  Cancel  all  factors  common  to  numerator  and  denominator. 

2.  When  numerator  and  denominator  involve  different  powers 
of  the  same  letter  (or  expression)  as  factors,  cancel  the  lower 
power  and  subtract  its  exponent  from  that  of  the  higher  power. 

3.  Give  the  quotient  the  -\-  or  —  sign,  according  as  the 
numerator  and  denominator  have  the  same  or  opposite  signs. 

Thus,        =  6a'^-2  =  ba^,  and  = = 

car  —a'  a' -2         a* 

322  Rules  for  dividing  by  a  monomial.     From  the  definition  of 
division  and  §  320,  4,  we  derive  the  following  rules. 

1.  To  divide  one  monomial  by  another,  form  a  fraction  by 
writing  the  dividend  over  the  divisor,  and  simplify. 


THE    FUNDAMENTAL   OPERATIONS  109 

2.  To  divide  a 'pol ynoviial  hy  a  monoviial,  divide  each  term  of 
the  dividend  by  the  divisor,  and  add  the  quotients  so  obtained. 

—  8  a^b'^c  4  a^c 

Thus,  —  8  a^b-c  -^  6  a¥'d  = = ,  by  cancelling  the 

Gab^d  '6¥d  ^ 

common  factor  2  ab-  and  applying  the  rule  of  signs. 

Again,       {az^  —  4  a-z-)  -^  ax  = =  x'  —  i  ax. 

ax         ax 

But  when  ci  lias  no  factor  in  common  with  a  and.  b,  we 
regard  {a-{-b)/d  as  a  simpler  form  of  the  quotient  than 
a/d  +  b/d. 

Division  of  a  polynomial  by  a  polynomial.     If  A  and  B  are     323 

polynomials  which  have  common  factors,  the  quotient  is 
the  expression  to  which  A/D  reduces  when  these  factors  are 
cancelled. 

Thus,  \i  A=x'^  —  y"-,  B  =  x--{-2xy  +  y",  the  quotient  is  {x  —  y)/(x  +  y). 
A  x2  -  2/2  (xj^y)(x-y)      x-y 


For 


B      x'^  +  -2xy  +  y-  (x  +  y)'^  x  +  y 


In  another  chapter  we  shall  give  methods  for  finding  the 
factors  which  are  common  to  two  polynomials.  The  process 
called,  long  division  is  considered  in  Chapter  V. 

Complex   expressions.      Observe    that    a  -i-  b  x  c   means    j  c,     324 
while  a  -=-  be,  like  a  -^  (b  x  c),  means  a  [be. 

In  the  chapter  on  fractions  we  shall  consider  complex 
expressions  in  which  a  number  of  indicated  multiplications 
and  divisions  occur.     In  particular  we  shall  find  that 

ax(hxc^d)=axbxc^d.  (1) 

a^{bxc-T-d)=a-^b-^cxd.  (2) 

In  (1)  the  signs  x  and  -f-  within  the  parentheses  remain 
unchanged  when  the  parentheses  are  removed ;  but  in  (2)  each 
X  is  changed  to  -i-,  and  each  -t-  to  x . 


110  A   COLLEGE   ALGEBRA 

EXERCISE  IV 

1.  Divide  15  a^bc-  by  10  ab^c'^ 

2.  Divide  75x^2"  by  -  100  ax^z^. 

3.  Divide  -  35x2'"^"  by  28x"'2/'"  +  ". 

4.  Divide  -  54  \  {ab^-)'^c  1 5  by  -  1 8  ^  a  (l^c)^  ]  ^ 

5.  Divide  x-y  —  xy-  by  x^  —  |/2_ 

6.  Divide  (x^  -  y^)  (x^  +  y^)  by  (x  -  y)  {x-  -  xy  +  y^). 
(a  -  6)2(6  _  c)3(c  -  a)* 


7.  Simplify 

8.  Simplify 

9.  Simplify 


(6  -  a)(c-  bY{a  -  c)3 

30  a263c4  -  25  a^ftScS  +  20  a*¥c^ 

3(x_2/)4_2(x-2/)3  +  5(x-2/)2 


(2/  -  xY 

10.  Simplify  4  a^  x  (3  a63c2)2  ^  (a6c)2  -  6  6c. 

11.  Simplify  the  following  (1)  by  performing  the  operations  In  the 
order  indicated,  (2)  by  first  removing  the  parentheses. 

a?  ^\a^  ^  (a*  ^  ofi  y.  a)  y.  (a^  x  a  ^  a"^)]. 

12.  By  what  must  2  a  {x-y^Y  ^e  multiplied  to  give  -  4  a2  {z^y'^Y  ? 

III.     SIMPLE    EQUATIONS    IN    ONE  UNKNOWN 
LETTER 

CONDITIONAL  EQUATIONS 

325  The  expressions  3  a;  —  4  and  a;  +  6  are  not  identically  equal, 
§  281,  and  therefore  are  not  equal  in  value  for  all  values  of  x. 
If  asked,  "  For  what  value  or  values  of  x  are  the  values  of 
these  expressions  the  same  ?  ",  we  begin  by  supposing  them 
to  be  the  same,  and  state  the  supposition  thus : 
3a;-4  =  a:  +  6. 
The  expression  thus  formed  is  called  a  conditional  equatio?i, 
or  an  equation  of  condition,  because  it  states  a  condition  which 


SIMPLE   EQUATIONS  111 

the  "  unknown  letter  "  x  is  to  satisfy.  It  serves  the  purpose 
of  restricting  x  to  values  which  satisfy  this  condition,  being 
true  when  the  values  of  3  a;  —  4  and  a;  +  6  are  the  same  and 
then  only. 

Similarly  a:  +  y  =  0  is  a  conditional  equation  in  the  two 
unknown  letters  x  and  y,  and,  in  general. 

When  the  exjjresslons  A  ayul  B  are  not   identically  equal,     326 
A  =  B  is  a  conditional  equation.      This  equation  means  :  ''  A  and 
B  are  sujj/josed  to  hai^e  equal  values.''     And  it  restricts  the 
variable  letters  in  A  and  B  to  values  for  tvhich  this  sujjposition 
is  true. 

The  letters  whose  values  the  equation  A  =  B  thus  restricts 
are  called  the  unknoxcn  letters  of  the  equation. 

In  what  follows,  the  word  "  equation "  will  mean  "  con- 
ditional equation." 

If  the  only  letters  in  an  equation  are  the  unknown  letters,     327 
as  X,  y,  z,  we  call  it  a  numerical  equation ;  but  if  there  are  also 
known  letters,  as  a,  h,  c,  we  call  it  a  literal  equation. 

Thus,  2x  —  Sy  =  5isa, numerical,  but  ax  +  by  =  c  is  a. literal  equation. 
A  literal  equation  does  not  restrict  the  values  of  the  known  letters. 

If  both  A  and  B  are  rational  and  integral  with  respect  to     328 
the  unknown  letters,  the  equation  .1  =  B  is  said  to  be  rational 
and  integral.     But  if  A   or  B  is  irrational  or  fractional,  the 
equation  is  said  to  be  irrational  or  fractional. 

No  account  is  taken  of  numbers  or  known  letters  in  this  classification. 
Thus,  V2x  +  y/b  —  c  is  both  rational  and  integral. 

In  the  case  of  a  rational  integral  equation  reduced  to  its     329 
simplest  form,  §  340,  the  degree  of  the  term  or  terms  of  highest 
degree  is  called  the  degree  of  the  equation  itself. 

Thus,  the  degree  of  ax'^  -\-hx  =  cis  two ;  that  of  xH^  -\-  y*  =  b  is  five. 
The  degree  is  measured  with  respect  to  all  the  unknown  letters,  but 
these  letters  only. 


112  A   COLLEGE    ALGEBRA 

330  Equations  of  the  first  degree  are  often  called  simple  or  linear 
equations  ;  those  of  the  second,  third,  fourth  degrees  are  called 
quadratic,  cubic,  biquadratic  equations  respectively. 

331  An  equation  in  one  unknown  letter,  as  x,  restricts  oc  to  a 
finite  number  of  values.  We  say  that  these  values  of  x  satisfy 
the  equation,  or  that  they  are  its  solutions  or  roots.     Hence 

332  A  root  of  an  equation  in  x  is  any  number  or  knoxcn  expression 
which,  if  substituted  for  a:,  will  make  the  equation  an  identity. 

Thus,  1  and  —  2  are  roots  of  the  equation  z^  ^  x  =  2  ;  for  12  +  1  =  2 
and  {-2)2  +  (-2)ee2. 

Again,  a  —  6  is  a  root  of  z  +  6  =  a  ;  for  (a  —  6)  -f  6  =  a. 

333  Notes.  1.  An  equation  may  have  no  root  ;  for  it  may  state  a  condi- 
tion which  no  number  can  satisfy. 

Thus,  no  finite  number  can  satisfy  the  equation  x  +  2  =  x  +  3. 

2.  In  every  equation  in  x  which  /i«s  roots,  x  is  merely  a  sym.bol  for  one 
or  other  of  these  roots.  In  fact  the  equation  itself  is  merely  a  disguised 
identity,  a  substitute  for  the  several  actual  identities  obtained  by  replacing 
X  by  each  root  in  turn. 

Thus,  z2  ^  X  =  2  is  merely  a  substitute  for  the  two  identities  12  +  1=2 
and  (-2)2  + (-2)  =  2. 

ON   SOLVING  EQUATIONS 

334  To  solve  an  equation  in  one  unknown  letter  is  to  find  all  its 
roots,  or  to  prove  that  it  has  no  root. 

The  reasoning  on  which  the  process  depends  is  illustrated 
in  the  following  examples. 

Example  1.     Solve  the  equation  3z  —  4  =  x  +  6. 

Starting  with  the  supposition  that  x  has  a  value  for  which  this  equation 
is  true,  we  may  reason  as  follows : 

If  3x-4  =  x  +  6, 

then  3x-4  +  (-x  +  4)  =  x  +  6  +  (-x  +  4), 

or  2x  =  10, 

and  therefore  x  =  5. 

Hence,  if  3x  —  4  =  x  +  6,  then  x  =  5. 


(1) 

(2) 

§249 

(3) 

§300 

(4) 

§253 

(a) 

SIMPLE    EQUATIONS  113 

The  proposition  (a)  thus  proved  states  that  if  (1)  is  ever  true,  it  is 
when  a:  =  5,  in  other  words,  that  the  only  number  which  can  be  a  root 
of  (1)  is  5;  but  it  does  not  state  tliat  5  is  a  root  of  (1).  Tliat  statement 
would  be 

If  X  =  5,  then  3  x  -  4  =  x  +  6.  (6) 

And  (6)  is  not  the  same  as  (a)  but  its  converse,  §  291. 

We  may  prove  that  5  is  a  root  of  (1)  by  substituting  5  for  x  in  (1) ; 
for  we  thus  obtain  the  true  identity  3-5  —  4  =  5  +  6. 

But  this  step  is  not  necessary,  except  to  verify  the  accuracy  of  our 
reckoning.  For  when  a  true  proposition  has  been  proved  by  a  reversible 
process,  we  may  always  conclude  that  its  converse  is  true,  §  293.  And 
this  is  the  case  with  (a),  since  the  process  by  which  (4)  was  derived 
from  (1)  is  made  up  of  reversible  steps  and  is  therefore  reversible  as  a 
whole.     Thus, 


If 

x  =  5, 

(4) 

then 

2x  =  10, 

(3) 

§253 

or                     3x  - 

-4  +  (-x  +  4)  =  x  +  G  +  (- 

■x  +  i), 

(2) 

§300 

and  therefore 

3  X  -  4  =  X  +  6. 

0) 

§249 

Hence,  in  proving  the  proposition  (o)  by  a  reversible  process,  we  have 
at  the  same  time  proved  the  converse  proposition  (6),  that  is,  we  have  not 
only  proved  that  no  other  number  than  5  can  be  a  root  of  (1),  but  also 
that  5  is  itself  a  root  of  (1). 

Example  2.     Solve  the  equation  x^  =  9. 

If  X-  =  9,  (1) 

then  either  x  =  3,       (2)  or  x  =  -  3.       (3)       §  257 

Hence  (1)  can  have  no  other  roots  than  3  and  —  3. 

But  both  3  and  —  3  are  roots  of  (1),  since  the  step  by  which  each  of 
the  equations  (2)  and  (3)  has  been  derived  from  (1)  is  reversible.  Thus,  (1) 
follows  from  (2)  and  also  from  (3),  §  257. 

These  examples  illustrate  the  following  general  principles: 

In  seeking  the  roots  of  an  egtiation  in  x,  we  treat  the  eqiia-     335 
tion  as  if  it  were  a  known  identity,  and  endeavor  to  find  all  the 
equations  of  the  form  x  =  c  which  necessarily  follow  from  it, 
ivhen  thus  regarded,  by  the  rules  of  reckoning. 

If  the  process  by  which  one  of  these  equations  x  =  c  has  been 
derived  is  reversible  when  x  has  the  value  c,  we  may  at  once 


114  A   COLLEGE   ALGEBRA 

conclude  that  c  is  a  root ;  and  the  process  is  reversible  if  it  is 
made  up  of  reversible  steps. 

336  It  is  important  to  remember  that  the  mere  fact  that  a 
certain  value  of  x  has  been  derived  from  an  equation  by  the 
rules  of  reckoning  does  not  prove  it  to  be  a  root.  The  process 
must  be  reversible  to  Avarrant  this  conclusion. 

Thus,  from  x  -  2  =  0,  (1) 

it  follows  that  (a;  -  2)  (x  -  3)  =  0,  (2)       §  253 

and  hence  that  either  x  =  2,  or  x  =  3.  (3)       §  253 

But  we  have  no  right  to  draw  the  absurd  conchision  that  3  is  a  root 
of  (1).  For  when  x  =  3  we  cannot  reverse  the  process,  that  is,  divide 
both  members  of  (2)  by  x  —  3,  since  the  divisor  x  —  3  is  then  0. 

On  the  other  hand,  when  x  =  2  we  can  reverse  tlie  process,  since  x  —  3 
is  then  not  0  but  —  1 ;  and  2  is  a  root  of  (1). 

TRANSFORMATION   THEOREMS 

337  In  the  light  of  what  has  just  been  said  we  may  regard  any 
correct  application  of  the  rules  of  reckoning  to  an  equation  as 
a  legitimate  transformation  of  the  equation  ;  and  if  such  a 
transformation  is  reversible,  we  may  conclude  that  it  leaves 
the  roots  of  the  equation  unchanged.  Hence  the  following 
theorems. 

338  Theorem  1.  The  following  transformations  of  an  equation 
leave  its  roots  unchanged,  namely : 

1.    Applying  the  rules  of  combination,  §  259,  to  each  member 


2.  Adding  any  expression  xvhich  has  a  finite  value  to  both 
members,  or  subtracting  it  from  both. 

3.  Multiplying  or  dividing  both  members  by  the  same  constant 
{7lOt  0). 

For  all  the  rules  of  reckoning  involved  in  these  transformations  are 
reversible,  §  259. 

We  may  also  state  the  proofs  of  2  and  3  as  follows : 


SIMPLE   EQUATIONS  115 

If  A  and  B  denote  expressions  in  a;,  the  roots  of  the  equation  A  =  3 
are  numbers  which  substituted  for  xin  A  and  B  make  A  =  B^%  332. 

But  any  value  of  z  which  makes  A^B  and  C  finite  will  make 
A  -y  C  =  B  -\-  C^  and  conversely,  §  249 ;  hence  the  roots  of  ^  =  i?  are 
the  same  as  those  oiA-\-C  —  B-\-C. 

Again,  if  c  denote  any  constant  except  0,  any  value  of  x  which  makes 
A^B  will  make  cA  =  cB,  and  conversely,  §  253 ;  hence  the  roots  oi  A  =  B 
are  the  same  as  those  of  cA  =  cB. 

Thus,  in  §  334,  Ex.  1,  the  equations 

3x-4  =  z  +  6,  (1) 

3x-4  +  (-x  +  4)  =  x  +  6  +  (-a;  +  4),  (2) 

2x  =  10,  (3) 

X  =  5.  (4) 

all  have  the  same  root,  5. 

Here  (2)  is  derived  from  (1)  by  the  transformation  2,  (3)  from  (2)  by 
the  transformation  1,  and  (4)  from  (3)  by  the  transformation  3. 

Corollary.      The  following    transformations   of  an    equation     339 
leave  its  roots  unchanged,  namely: 

1.  Transposing  a  term,  with  its  sign  changed,  from  one 
member  to  the  other. 

2.  Cancelling  any  terms  that  may  occur  hi  both  7nembers. 

3.  Changing  the  signs  of  all  terms  in  both  members. 

For  3  is  equivalent  to  multiplying  both  members  by  —  1.  And  1  and 
2  are  equivalent  to  subtracting  the  term  in  question  from  both  members 
of  the  equation. 

Thus,  if  from  both  members  oix  —  a-\-b=c+b  (1) 

we  subtract  —  a  +  b  =  —  a  +  h 

we  obtain  x  =      c  +  a.  (2) 

The  effect  of  the  subtraction  is  to  cancel  b  in  both  members  of  (1)  and 
to  transpose  —  a,  with  its  sign  changed,  from  the  first  member  to  the 
second. 

By  aid  of  these  transformations,  §§  338,  339,  every  rational     340 
integral  equation  in  a-  may,  without  changing  its  roots,  be 
reduced,  to  the  standard  form 

UqX"  +  ttio;""^  +  ■  •  •  +  «„-i^  +  «„  =  0. 


116  A    COLLEGE    ALGEBRA 

We  suppose  such  an  equation  reduced  to  this  form  when 
its  degree  is  measured,  §  329.  The  like  is  true  of  rational 
integral  equations  in  more  than  one  unknown  letter. 

Thus,  x2  +  3x  +  5  =  x2  —  4x  +  7  can  be  reduced  to  the  form  7  x  —  2  =  0. 
Its  degree  is  therefore  one,  not  two. 

341  Theorem  2.     Whe?i  A,  B,  aiid  C  are  integral,  the  equation 

AC  =  BC 
has  the  same  roots  as  the  two  equations 

A  =  B  and  C  =  0. 

For  any  value  of  x  which  makes  AC  =  BC  must  make  either  A  =  B 
or  C  =  0  ;  and,  conversely,  any  value  of  x  which  makes  either  ^  =  i.'  or 
C  =  0  will  make  ^C=BC,  §§251,  253. 

In  this  proof  it  is  assumed  that  A,  B,  C  have  finite  values  for  the 
values  of  x  in  question.  This  is  always  true  when,  as  is  here  supposed, 
A,  B,  C  are  integral  ;  but  it  is  not  always  true  when  A,  B,  C  are 
fractional. 

In  particular,  ivhen  A  and  C  are  integral,  the  equation  AC  =  0  has  the 
same  roots  as  the  equations  A  =  0  and  C  =  0  jointly. 

Thus,  the  roots  of  the  equation  x^  :=  3  x  are  the  same  as  those  of  the 
two  equations  x  =  3  and  x  =  0,  that  is,  3  and  0. 

Similarly  the  roots  of  (x  —  1)  (x  —  2)  =  0  are  the  same  as  those  of  the 
two  equations  x  —  1  =  0  and  x  —  2  =  0,  that  is,  1  and  2. 

342  Hence  the  effect  of  multiplying  both  members  of  an  inte- 
gral equation  A  =  B  hy  the  same  integral  function  C  is  to 
introduce  extraneous ^voots,  namely,  the  roots  of  the  equation 
C  =  0.  Conversely,  the  effect  of  removing  the  same  integral 
factor  C  from  both  members  of  an  integral  equation  AC  =  BC, 
is  to  lose  certain  of  its  roots,  namely,  the  roots  of  C  =  0. 

On  the  other  hand,  in  &  fractional  equation,  it  is  usually  the  case  that 
no  extraneous  roots  are  introduced  when  both  members  are  multiplied 
by  the  lowest  common  denominator  of  all  the  fractions. 

Thus,  if  the  equation  be  l/x=  l/(2x  —  1),  and  we  multiply  both 
members  by  x(2x  —  1),  we  obtain  2x  —  1  =  x,  whose  root  is  1.  As  1  is 
not  a  root  of  x(2x  —  1)  =  0,  we  have  introduced  no  extraneous  root. 


SIMPLE    EQUATIONS  117 

Corollary.      The  integral  equation  A^  =  B^  has  the  same  roots     343 
as  the  equations  A  =  B  and  A  =  —  H  jointly. 

For  A'^  =  B^  has  the  same  roots  as  A^  —  B-  =  0,  §  339.  And  since 
A^  -B'  =  {A-  B)  (J.  +  B),  the  equation  A'^  -  B- =  0  has  the  same  roots 
as  the  two  equations  A  —  B  —  0  and  A  +  B  =  0,  %Ml,  and  therefore  the 
same  roots  as  the  two  equations  A  —  B  and  A  =  —  B,  §  339. 

Thus,  the  roots  of  the  equation  (2x  —  1)2  =  (x  —  2)2  are  the  same  as 
those  of  the  two  equations  2x  —  l=x  —  2  and  2  x  —  1  =  —  (x  —  2),  that 
is,  —  1  and  1. 

Hence  the  effect  of  squaring  both  members  of  the  equation     344 
A  =  B  is  to  introduce  extraneous  roots,  namely,  the  roots  of 
the  equation  A  =  —  B.     Conversely,  the  effect  of  deriving  from 
A"^  =  B^  the  single  equation  A  —  B  is  to  lose  certain  of  the 
roots,  namely,  the  roots  of  the  equation  A  =  —  B. 

Since  A"  -  £"  =  (A  -  B)  (^"-^  +  A^'-'B  H \-B^-^),%  308,     345 

it  follows  by  the  reasoning  of  §  343  that  the  roots  of  .4"  =  ii" 
are  those  of  A  =  B  and  A"-^  +  A"--B  H \-  B"-^  =  0  jointly. 

Thus,  since  x^  —  1  =  (x  —  1)  (x2  +  x  +  1),  the  equation  x^  =  1  has  the 
same  roots  as  the  equations  x  =  1  and  x-  +  x  +  1  =  0  jointly. 

The  theorems  just  demonstrated,  §§  338-345,  hold  good  for     346 
equations  in  more  than  one  unknown  letter  if  the  word  root  be 
replaced  by  the  word  solution,  §  355. 

Thus,  by  §  339,  the  equation  x  +  2?/  —  3  =  0  (1)  has  the  same  solu- 
tions as  the  equation  x  =  —  2y  +  3  (2),  that  is,  every  pair  of  values  of 
X  and  y  which  satisfy  (1)  will  also  satisfy  (2),  and  conversely. 

Equivalent  equations.     When  two  or  more  equations  have  the     347 
same  roots  (or  solutions),  we  say  that  they  are  equivalent. 

Thus,  §  3§§,  the  equations  A  =  B  and  A  +  C  =  B  +  C  zxe.  equivalent. 
Again,  §341,  the  equation  AC  =  BC  is  equivalent  to  the  two  equations 
^  =  B  and  C  =  0. 

But_x?^  =  9  (1}  and  x  =  3  (2)  are  not  equivalent  although  both  have  the 
root  3.     For  (1)  also  has  the  root  — _3,  which  (2)  does  not  have. 


118  A   COLLEGE    ALGEBRA 


SOLUTION   OF  SIMPLE  EQUATIONS 

348  From  the  transformation  theorems  of  §§  338,  339  we  may 
at  once  derive  the  following  rule  for  solving  a  simple  equation 
in  one  unknown  letter,  as  x. 

To  solve  a  simple  equation  in  x,  reduce  it  to  the  form  ax  =  b, 
Then 

1.  If  a.=f^  0,  the  equation  has  the  single  root  b/a. 

2.  7/"  a  =  0,  and  b  g^  0,  the  equation  has  no  root. 

3.  i/"  a  =  0,  and  b  =  0,  the  equation  is  an  identity. 

If  the  equation  has  fractional  coefficients,  it  is  usually  best 
to  begin  by  multiplying  both  members  by  the  lowest  common 
denominator  of  these  fractions.  This  process  is  called  clearing 
the  equation  of  fractions. 

We  then  reduce  the  equation  to  the  form  ax  =  b  hj  trans- 
posing the  unknown  terms  to  the  first  member  and  the  known 
terms  to  the  second,  and  collecting  the  terms  in  each  member. 
To  verify  the  result,  substitute  it  for  x  in  the  given  equation, 

2  'r  X  ■ 2  X 

Example  1.     Solve  — —  =  -  -  (4  -  x). 

To  clear  of  fractions,  multiply  both  members  by  the  Led.,  6. 

Then  4x  -  3(x  -  2)  =  z  -  6(4  -  x), 

or  4x-3x  +  6:=x-24  +  6x. 

Transpose  and  collect  terms,  —  6  x  =  —  30. 

Therefore  x  =  5. 

■r.    ..     s-                     2-5      5-2      5,,^, 
Verification.  — —  =  -  —  (4  —  5). 

Example  2.     Sol-'e  mx  +  n  =  px -\-  q. 
Transpose  and  collect  terms,  [m  —  p)x  =  q  -  n. 
Hence  if  m  :^p,  the  equation  has  the  single  root  {q  —  n)/{m  —  p). 
If  m  =p  and  q  7^  n,  it  has  no  root. 

If  m  =  p  and  q  —  n,   it  is  an  identity   and  every  value  of  X 

satisfies  it. 


SIMPLE    EQUATIONS  119 


Example  3.     Solve     (x  +  a)(x  +  b)  =  {x  -  af. 
Expand,  x^  +  (a  +  6)  x  +  a6  =  x-  -  2  ax  +  a^. 

Cancel  x^,  and  transpose  and  collect  terms. 
Then  (3  a  +  6)x  =  a2  -  a6, 

a2-a6 


and  therefore 


3a  +  6 


Sometimes  a  root  of  an  equation  can  be  fomid  by  inspectioji.     349 
The  equation  is  then  completely  solved  if  it  be  a  simple  equa- 
tion, for  it  can  have  no  other  root  than  the  one  thus  found. 

Example.     Solve  (x  -  a)2  -  (x  -  6)2  =  (a  -  b)^ 

Evidently  this  is  a  simple  equation,  and  when  x  =  6  it  reduces  to  the 
identity  (6  —  a)2  =  (a  —  6)2.     Hence  its  root  is  b. 

The  roots  of  an  equation  of  the  form  AB  =  0,  in  which  A  350 
and  B  denote  integral  expressions  of  the  first  degree  in  x,  can 
be  found  by  solving  the  two  simple  equations  ^  =  0  and 
B  =  0,  §  341.  In  like  manner,  when  A,  B,  C  are  of  the  first 
degree,  the  roots  of  ABC  =  0,  AC  =  BC  and  A'^  —  B^  may 
be  found  by  solving  simple  equations,  §§  341,  343. 

Example  1.     Solve  (x  -  2)  (x  +  3)  (2  x  -  5)  (3  x  +  2)  =  0. 
This  equation  is  equivalent,  §  347,  to  the  four  equations 

x-2  =  0,  x  +  3  =  0,  2x-5  =  0,  3x  +  2  =  0. 
Hence  its  roots  are  2,  -  3,  5/2,  -  2/3. 
Example  2.     Solve  4  x2  -  5  x  =  3  x2  +  7  x. 
This  equation  has  the  same  roots  as  the  two  equations 

X  =  0  and  4x  —  5  =  3x  +  7. 
Its  roots  are  therefore  0  and  12. 

EXERCISE  V 
Solve  the  following  equations. 

1.  15- (7-5x)  =  2x  +  (5-3x). 

2.  X  (x  +  3)  -  4  X  (X  -  5)  =  3  X  (5  -  x)  -  16. 

3.  (X  +  1)  (X  -f  2)  -■  (X  4-  3)  (X  +  4)  =  0. 


120  A   COLLEGE    ALGEBRA 


X        X        X         X 

'  +  2  +  4  +  8  +  1^ 


5.  z-2[x-3(x  +  4)-5]  =  3^2x-[x-8(x-4)]^  -2. 

6.  2^3[4{5x-l)-8]-20^ -7  =  1. 

7.  Ui[Hi*-l)-6]  +  4^=l- 

5X-.4      1.3 X  -  .05  _  13.95  -  8x 
^^  'J~^  2  ~         1.2 

11.  3  ex  -  5  a  +  6  -  2  c  =  6  6  -  (a  +  3  6x  +  2  c). 

12.  (6  -  c)  (a  -  X)  +  (c  -  a)  (6  -  X)  +  (a  -  6)  (c  -  X)  =  1  -  x. 

13. [-  -  =  2,  by  inspection. 

a  +  1      a 


1  A 

x  +  1 
a  +  6 

X  — 

+  — 
a  - 

1   _ 

2a 

a^  -  6-^ 

15. 

m 

n 

2x. 

16.  (2x-l)(3x-l)(4x  +  l)(5x  +  2)  =  0. 

17.  (x2  -  x)  (2  X  -  5)  =  (x2  -  x)  (x  +  9). 

18.  (X  +  2)3  -  (X  -  2)3  =  32  X  +  IG. 

19.  [(a  +  6)x  -  c]-^  =  [(a  -  6)x  +  c]2. 

20.  (x2  -  2  X  +  1)2  -  (x  -  1)2  (X  -  3)2  =  0. 

PROBLEMS 

351  On  solving  problems.  In  the  following  problems  it  is  required 
to  derive  the  values  of  certain  unknown  numbers  from  given 
relations,  called  the  conditions  of  the  j)roblem,  connecting  these 
numbers  with  known  numbers  and  one  another. 

In  each  case  we  represent  one  of  the  unkno^vn  numbers  by 
a  letter,  as  x.     The  given  conditions  then  enable  us  to  express 


SIMPLE    EQUATIONS  121 

the  remaining  unknown  numbers  in  terms  of  x  and  to  form  a 
single  equation  connecting  the  expressions  thus  obtained.  This 
equation  is  the  statement  of  the  problem  in  algebraic  symbols. 
We  solve  it  for  x.  If  the  problem  have  any  solution,  it  will 
be  the  value  thus  found  for  x,  together  with  the  corresponding 
values  of  the  other  vmknown  numbers. 

It  may  happen,  however,  that  the  value  thus  found  for  x  is 
not  an  admissible  solution  of  the  problem.  For  the  problem 
may  be  one  which  imposes  a  restriction  on  the  character  of 
the  unknown  numbers,  as  that  they  be  integers,  and  the  equa- 
tion in  X  into  which  the  statement  of  the  problem  has  been 
translated  does  not  express  this  restriction. 

Having  solved  the  equation  in  x,  therefore,  we  must  notice 
whether  the  result  is  a  number  of  the  kind  reqiiired  before  we 
accept  it  as  a  solution  of  the  problem.  If  it  is  not,  we  conclude 
that  the  problem  is  an  impossible  one. 

Example  1.  The  sum  of  the  digits  of  a  certain  number  of  two  digits 
is  12.  If  we  reverse  the  order  of  the  digits  we  obtain  a  number  which  is 
4/7  as  great.     What  is  the  number  ? 

Here  there  are  four  unknown  numbers,  namely,  the  tens  digit,  the 
units  digit,  the  value  of  the  number  as  it  stands,  and  the  value  when  the 
digits  are  reversed ;  but  all  four  can  be  readily  expressed  in  terms  of 
either  units  or  tens  digit. 

Thus,  let  X  =  the  tens  digit. 

Then  12  —  x  =  the  units  digit, 

lOx  +  (12  —  x)  =  value  of  required  number, 
10  (12  —  x)  +  X  =  value  with  digits  reversed. 

By  the  remaining  condition  of  the  problem,  we  have 

10(12  -  X) +  x  =  K10«  + (12 -X)].  (1) 

Solving  this  equation  we  obtain  x  =  8,  which  being  an  integer  less 
than  10,  is  an  admissible  solution  of  the  problem.  The  like  is  true  of 
12  —  X  or  4.     Hence  the  required  number  is  84. 

Notice  that  with  a  slight  modification  the  problem  becomes  impossible. 
Thus,  if  we  require  that  reversing  the  digits  shall  double  the  value  of  the 
number,  we  have,  instead  of  (1),  the  equation 

10(12 -x)  +  x  =  2[10x +  (12- X)].  (2) 


122  A   COLLEGE    ALGEBRA 

And  solving  (2)  we  obtain  x  =  32/9,  which  being  fractional  is  not  an 
admissible  solution  of  the  problem. 

"When  dealing  with  a  problem  which  has  to  do  with  certain 
magnitudes,  as  intervals  of  time,  remember  that  the  letters 
used  in  stating  the  problem  algebraically  are  to  represent 
not  the  magnitudes  themselves,  but  the  numbers  which  are 
their  measures  in  terms  of  some  given  unit  or  units.  Care 
must  also  be  taken  to  express  the  measures  of  all  magnitudes 
of  the  same  kind,  whether  known  or  unknown,  In  terms  of  the 
same  unit. 

Example  1.  A  tank  has  a  supply  pipe  A  which  will  fill  it  in  3  hours, 
and  a  waste  pipe  B  which  will  emiJty  it  in  3  hours  and  40  minutes.  K 
the  tank  be  empty  when  both  pipes  are  opened,  how  long  will  it  be  before 
the  tank  is  full  ? 

Let  X  denote  the  number  of  hours  required. 

Then  1  /x  is  the  part  filled  in  one  hour  when  both  A  and  B  are  open. 

But  were  A  alone  open,  the  part  filled  in  one  hour  would  be  1/3. 

And  were  B  alone  open  (and  water  in  the  tank)  the  part  emptied  in 
one  hour  would  be  l/3j  or  3/11. 

XT  113  2 

Hence  -  = ,  or  — 

X      3      11         33 

Therefore  x  =  33/2  hours,  or  16  hours  30  minutes. 

Example  2.  A  crew  can  row  2  miles  against  the  current  in  a  certain 
river  in  15  minutes ;  with  the  current  in  10  minutes.  What  is  the  rate  of 
the  current  ?     And  at  what  rate  can  the  crew  row  in  dead  water  ? 

Let  X  =  rate  of  current  in  miles  per  minute. 

As  the  rate  of  the  crew  against  the  current  is  2/15  in  miles  per 
minute,  in  dead  water  it  would  be  2/15  +  x. 

And  as  the  rate  of  the  crew  with  the  current  is  2/10.  or  1/5  in  miles 
per  minute,  in  dead  water  it  would  be  1  /5  —  x. 

Hence  VV  +  ^  =  i  ~  s;, 

whence  *  =  t^  (miles  per  minute), 

and  .j^  4-  X  =  ^  (miles  per  minute). 

Example  3.     At  what  time  between  two  and  three  o'clock  do  the  hour 
and  minute  hands  of  a  clock  point  in  opposite  directions  ? 
Let  X  =  number  of  minutes  past  two  o'clock  at  the  time  required. 


SBIPLE   EQUATIONS  123 

Since  the  minute  hand  starts  at  XII  it  will  then  have  traversed  x 
minute  spaces. 

The  hour  hand  starts  at  II,  or  10  minute  spaces  in  advance  of  the 
minute  hand,  but  it  moves  only  1/12  as  fast  as  the  minute  hand. 

Therefore  when  the  minute  hand  is  at  x  minute  spaces  past  XII,  the 
hour  hand  is  at  10  +  x/12  minute  spaces  past  XII. 

But  by  the  conditions  of  the  problem,  at  the  time  required  the  minute 
hand  is  30  minute  spaces  in  advance  of  the  hour  hand. 

Hence  x  =  Ao  +  — )  +  30, 

or  solving,  x  =  43  J'y  minute  spaces. 

Therefore  the  hands  point  in  opposite  directions  at  43j7j-  minutes  after 
two  o'clock,  or  16/j-  minutes  before  three  o'clock. 

Sometimes  in  the  statement  of  a  problem  the  known  num-  354 
bers  are  denoted  by  letters,  as  o,  b,  c.  The  value  found  for  x 
will  then  be  an  expression  in  a,  b,  c  which  may  represent  an 
admissible  solution  of  the  problem  for  certain  values  of  these 
letters,  but  not  for  others.  The  discussion  of  the  follow- 
ing problem,  known  as  the  problem  of  coiiriers,  will  illustrate 
this  point. 

Example.  Two  couriers  A  and  B  are  traveling  along  the  same  road 
in  the  same  direction  at  the  rates  of  m  and  n  miles  an  hour  respectively. 
B  is  now  d  miles  in  advance  of  A.  Will  they  ever  be  together,  and  if 
so,  when  ? 

Let  X  =  the  number  of  hours  hence  when  they  will  be  together. 

A  will  then  have  traveled  mx  miles,  and  B  nx  miles ;  and  since  B  is 
now  d  miles  in  advance  of  A,  we  have 


mx  =  nx  +  d. 

(1) 

whence 

{m  —  n)x  =  d. 

(2) 

and  therefore 

X  = hours  hence. 

(3) 

1.  If  A  is  to  overtake  B,  this  value  of  x  must  be  positive  ;  and  since 
by  hypothesis  d,  m,  n  all  denote  positive  numbers,  this  requires  that 
711  >  n.  Which  corresponds  to  the  obvious  fact  that  if  A  is  to  overtake 
B,  he  nmst  travel  faster  than  B  does. 

2.  At  the  same  time  we  can  interpret  the  negative  value  which  x  takes 
if  we  suppose  m  <  n  as  meaning  that  A  and  B  vtere  together  d/(n  —  m) 
hours  ago. 


124  A   COLLEGE   ALGEBRA 

3.  If  m  =  n,  we  cannot,  properly  speaking,  derive  (3)  from  (1),  since 
the  process  involves  dividing  by  in  —  n  which  is  0.  But  we  can  derive 
(3)  from  (1)  if  m  differs  at  all  from  ?i,  it  matters  not  how  little.  And  if 
in  (3)  we  regard  m  as  a  variable,  which  while  greater  than  n  is  continually 
approaching  equality  with  n,  the  fraction  d/{m  —  n)  becomes  a  variable 
which  continually  increases,  and  that  without  limit,  §  510.  All  of  which 
corresponds  to  the  obvious  fact  that  the  smaller  the  excess  of  A's  rate 
over  B's,  the  longer  it  will  take  A  to  overtake  B,  and  that  A  will  jiever 
overtake  B  if  his  rate  be  the  same  as  B's  rate. 

4.  Finally,  if  we  suppose  both  m  =  n  and  d  =  0,  the  equation  (1)  is 
satisfied  by  every  value  of  x.  Which  corresponds  to  the  obvious  fact  that 
if  A  and  B  are  traveling  at  the  same  rate  and  are  now  together,  they 
■will  always  be  together. 

EXERCISE  VI 

1.  The  sum  of  the  digits  of  a  certain  number  of  two  digits  is  14.  If 
the  order  of  the  digits  be  reversed,  the  number  is  increased  by  18. 
What  is  the  number  ? 

2.  By  what  number  must  156  be  divided  to  give  the  quotient  11  and 
the  remainder  2  ? 

3.  There  are  two  numbers  whose  difference  is  298.  And  if  the  greater 
be  divided  by  the  less,  the  quotient  and  remainder  are  both  12.  What 
are  the  numbers  ? 

4.  The  tens  digit  of  a  certain  number  of  two  digits  is  twice  the  units 
digit.  And  if  1  be  added  to  the  tens  digit  and  5  to  the  units  digit,  the 
number  obtained  is  three  times  as  great  as  if  the  order  of  the  digits  be 
first  reversed  and  then  1  be  subtracted  from  the  tens  digit  and  5  from 
the  units  digit.     What  is  the  number  ? 

5.  If  2  be  subtracted  from  a  certain  number  and  the  remainder  be 
multiplied  by  4,  the  same  result  is  obtained  as  if  twice  the  number  and 
half  a  number  one  less  be  added  together.     What  is  the  number  ? 

6.  A  father  is  now  four  times  as  old  as  his  son.  If  both  he  and  his 
son  live  20  years  longer,  he  will  then  be  twice  as  old  as  his  son.  What  are 
the  present  ages  of  father  and  son,  and  how  many  years  hence  will  the 
father  be  three  times  as  old  as  the  son  ? 

7.  A  tank  can  be  filled  by  one  pipe  in  3  hours,  and  emptied  by  a  second 
in  2  hours,  and  by  a  third  in  4  hours.  How  long  will  it  take  to  empty  the 
tank  if  it  start  full  and  all  the  pipes  are  opened  ? 


SIMPLE   EQUATIONS  125 

S.  A  and  B  can  do  a  certain  piece  of  work  in  10  days  ;  but  at  the  end 
of  the  seventh  day  A  falls  sick  and  B  finishes  the  piece  by  working 
alone  for  5  days.  How  long  would  it  take  each  man  to  do  the  entire 
piece,  working  alone  ? 

9.  At  what  time  between  eight  and  nine  o'clock  do  the  hands  of  a 
watch  point  in  the  same  direction  ?  in  opposite  directions  ? 

10.  How  soon  after  four  o'clock  are  the  hands  of  a  watch  at  right 
angles  ? 

11.  In  a  clock  which  is  not  keeping  true  time  it  is  observed  that  the 
interval  between  the  successive  coincidences  of  the  hour  and  minute  hands 
is  66  minutes.     What  is  the  error  of  tlie  clock  (in  seconds  per  hour)  ? 

12.  Four  persons,  A,  B,  C,  D,  divide  $1300  so  that  B  receives  f  as 
much  as  A,  C  f  as  much  as  B,  and  D  f  as  much  as  C.  How  much 
does  each  receive  ? 

13.  A  man  leaves  J  his  property  and  .f  1000  besides  to  his  oldest  son  ; 
I  of  the  remainder  and  $1000  besides  to  his  second  son  ;  h  of  the  sum 
still  remaining  and  $1000  besides  to  his  youngest  son.  If  $3500  still 
remain,  what  is  the  amount  of  the  entire  property  ? 

^^    14.    If  2  feet  be  added  to  both  sides  of  a  certain  square,  its  area  is 
increased  by  100  square  feet.     What  is  the  area  of  the  square  ? 

15.  The  height  of  a  certain  flagstaff  is  unknown  ;  but  it  is  observed 
that  a  flag  rope  fastened  to  the  top  of  the  staff  is  2  feet  longer  than  the 
staff,  and  that  its  end  just  reaches  the  ground  when  carried  to  a  point  18 
feet  distant  from  the  foot  of  the  staff.     What  is  the  height  of  the  staff  ? 

16.  A  purse  contains  a  certain  number  of  dollar  pieces,  twice  as  many 
half-dollar  pieces,  and  three  times  as  many  dimes.  If  the  total  value  of 
tlie  pieces  is  $11.50,  how  many  pieces  are  there  of  each  kind  ? 

17.  A  man  invests  $5000,  partly  at  6%  and  partly  at,  4%,  so  that  the 
average  rate  of  interest  on  the  entire  investment  is  [>]%.  What  sum 
does  he  invest  at  each  rate  ? 

18.  In  what  proportions  should  two  kinds  of  coffee  worth  20  cts.  and 
30  cts.  a  pound  respectively  be  combined  to  obtain  a  mixture  worth 
26  cts.  a  pound  ? 

19.  A  pound  of  a  certain  alloy  of  silver  and  copper  contains  2  parts  of 
silver  to  3  of  copper.  How  much  copper  must  be  melted  with  this  alloy 
to  obtain  one  which  contains  3  parts  of  silver  to  7  of  copper  ? 


126  A   COLLEGE   ALGEBRA 

20.  If  a  certain  quantity  of  water  be  added  to  a  gallon  of  a  given 
liquid,  it  contains  30%  of  alcohol ;  if  twice  this  quantity  of  water  be 
added,  it  contains  20%  of  alcohol.  How  much  water  is  added  each  time, 
and  what  percentage  of  alcohol  did  the  original  liquid  contain  ? 

21.  A  train  whose  rate  of  motion  is  45  miles  per  hour  starts  on  its  trip 
from  Philadelphia  to  Jersey  City  at  10  a.m.,  and  at  10.30  a.m.  another 
train  whose  rate  is  50  miles  an  hour  starts  on  its  trip  from  Jersey  City  to 
Philadelphia.  Assuming  that  the  two  cities  are  90  miles  apart,  when 
will  the  trains  pass  each  other,  and  at  what  distance  from  Jersey  City  ? 

22.  If  two  trains  start  at  the  times  mentioned  in  the  preceding  exam- 
ple and  pass  each  other  at  a  point  half  way  between  Jersey  City  and 
Philadelphia,  and  if  the  slower  train  moves  |  as  fast  as  the  swifter  one, 
what  are  their  rates,  and  when  do  they  pass  each  other  ? 

23.  A  rabbit  is  now  a  distance  equal  to  50  of  her  leaps  ahead  of  a  fox 
which  is  pursuing  her.  How  many  leaps  will  the  rabbit  take  before  the 
fox  overtakes  her  if  she  takes  5  leaps  while  the  fox  takes  4,  but  2  of 
the  fox's  leaps  are  equivalent  to  3  of  her  leaps  ? 

24.  If  19  ounces  of  gold  weigh  but  18  ounces  when  submerged  in 
water,  and  10  ounces  of  silver  then  weigh  9  ounces,  how  many  ounces 
of  silver  and  of  gold  are  there  in  a  mass  of  an  alloy  of  the  two  metals 
which  weighs  387  ounces  in  air  and  351  ounces  in  water  ? 

25.  A  traveler  set  out  on  a  journey  with  a  certain  sum  of  money  in 
his  pocket  and  each  day  spent  \  of  what  he  began  the  day  with  and  $2 
besides.  At  the  end  of  the  third  day  his  money  was  exhausted.  How 
much  had  he  at  the  outset  ? 

26.  The  base  of  a  certain  pyramid  is  a  square,  and  the  altitude  of  each 
of  the  triangles  which  bound  it  laterally  is  equal  to  an  edge  of  the  base. 
Were  this  edge  and  altitude  each  increased  by  3  inches,  the  area  of  the 
pyramid  would  be  increased  by  117  square  inches.  What  is  the  area  of 
the  pyramid  ? 

27.  The  sum  of  the  digits  of  a  certain  number  of  two  digits  is  a.  If 
the  order  of  the  digits  be  reversed,  the  number  is  increased  by  b.  What 
is  the  number?  Show  that  the  solution  is  admissible  only  when  9a>b 
and  when  both  9 a  +  6  and  9a  —  b  are  exactly  divisible  by  18. 

28.  Two  persons  A  and  B  are  now  a  and  6  years  old  respectively.  Is 
there  a  time  when  A  was  or  when  A  will  be  c  times  as  old  as  B,  and  if 
so,  when  ? 

Discuss  the  result  for  various  values  of  a,  b,  c,  as  iu  §  354. 


SIMULTANEOUS    SIMPLE    EQUATIONS  127 

IV.     SYSTEMS    OF   SIMULTANEOUS    SIMPLE 
EQUATIONS 

SIMULTANEOUS   EQUATIONS 

A  conditional  equation  in  two  unknown  letters,  as  x  and  y,     355 
will  be  satisfied  by  infinitely  many  pairs  of  values  of  these 
letters.     We  call  every  such  pair  a  solution  of  the  equation. 
The  like  is  true  of  an  equation  in  more  than  two  unknown 
letters. 

Thus,  the  equation  2  x  +  y  =  3  (1)  is  satisfied  if  we  give  any  value  what- 
soever to  X  and  tlie  corresponding  value  of  3  —  2x  to  y.  For  in  (1) 
substitute  any  number,  as  b,  for  x  and  3  —  26  for  y,  and  we  have  the  true 
identity  2  6  +  (3  -  2  6)  =  3. 

Thus,  x  =  0,  ?/  =  3;x  =  l,  2/  =  l;x  =  2,  ?/  =  — 1;  •••are  solutions 
of  (1). 

Note.     When  two  unknown  letters,  x,  y,  are  under  consideration,  the     356 
equation  x  =  2  means  that  x  is  to  have  the  value  2,  and  y  any  value  what- 
soever; in  other  words,  the  equation  x  =  2  then  has  an  infinite  number 
of  solutions.     And  the  like  is  true  of  any  equation  which  involves  but 
one  of  the  unknown  letters. 

It  is  therefore  natural  to  inquire  whether  there  may  not  be     357 
pairs  of  values  of  x  and  y  which  will  satisfy  two  given  equa- 
tions in  these  letters.     Such  pairs  usually  exist. 

Thus,  both  the  equations  2  x  +  ?/  =  3  and  4  x  +  3  ?/  =  5  are  satisfied 
when  X  =  2  and  2/  =  -  1 ;  for  2  ■  2  +  (-  1)  =  3,  and  4  •  2  +  3(-  1)  =  5. 

Simultaneous  equations.     Two  or  more  equations  involving     358 
certain  unknown  letters  are  said  to  be  siynultaneous  when  each 
unknown  letter  is  supposed  to  stand  for  the  same  number  in 
all  the  equations. 

Thus,  the  equations  2x  +  i/  =  3  (1)  and  4x  +  3?/  =  5  (2)  are  simul- 
taneous if  we  suppose  x  to  denote  the  same  number  in  (1)  as  in  (2),  and 
y  similarly. 

It  is  not  necessary  that  all  the  unknown  letters  occur  in  every  one  of  the 
equations.  Thus,  x  =  2,  y  =  3  constitute  a  pair  of  simultaneous  equations 
in  X  and  y. 


128  A   COLLEGE    ALGEBRA 

359  Generally  speaking,  the  supposition  that  certain  equations 
are  simultaneous  is  allowable  only  when  the  number  of 
equations  is  equal  to,  or  less  than,  the  number  of  unknown 
letters. 

Thus,  the  two  equations  cc  =  2  and  x  =  3  cannot  be  simultaneous, 
since  x  must  denote  different  numbers  in  the  two. 

360  A  solution  of  a  system  of  simultaneous  equations  is  any  set 
of  values  of  the  unknown  letters  which  will  satisfy  all  the 
equations  of  the  system. 

Thus,  x  =  2,  ?/  =  -lisa  solution  of  the  system 

2x  +  2/  =  3,  4x  +  3t/  =  5. 

361  To  solve  a  system  of  simultaneous  equations  is  to  find  all  its 
solutions  or  to  prove  that  it  has  no  solution. 

362  The  reasoning  on  which  the  process  depends  is  similar  to 
that  described  and  illustrated  in  §§  334,  335. 

Thus,  in  the  case  of  a  pair  of  equations  in  x  and  y  we  begin 
by  supposing  that  x  and  y  actually  have  values  which  satisfy 
both  equations.  On  this  supposition  the  equations  may  be 
treated  like  identities  and  the  rules  of  reckoning  applied  to 
them.  By  aid  of  these  rules  w^e  endeavor  to  transform  the 
equations  into  one  or  more  pairs  of  equations  of  the  form 
X  =  a,  y  =  b.  If  the  process  by  which  such  a  pair  x  =  a, 
y  =b  has  been  derived  is  reversible  when  x,  y  have  the  values 
a,  b,  we  may  at  once  conclude  that  a,  b  is  one  of  the  solu- 
tions sought;  and  the  process  is  reversible  if  it  consists  of 
reversible  steps. 

The  only  new  principle  involved  in  all  this  is  the  following: 

363  Principle  of  substitution.  Jf  from,  the  supposition  that  all  the 
given  equations  are  actually  satisfied  it  follows  that  the  values 
of  a  certain  pair  of  expressions,  A  and  B,  are  the  same,  the 
one  expression  may  be  substituted  for  the  other  in  any  of  the 
equatiotis. 


SIMULTANEOUS    SIMPLE    EQUATIONS  129 

Example.     Solve  the  pair  2x  +  y  =  12,  (1) 

2/  =  8.  (2) 

From  the  supposition  that  x  and  y  actually  have  values  v?hich  satisfy 
both  equations  it  follows  that  the  value  of  ij  in  (2)  and  therefore  in  (1)  is  8. 
Substituting  this  value,  8,  for  y  in  (1),  we  obtain 

2a;+8^12,  (3) 

whence  x  =  2.  (4) 

Therefore,  if  (1),  (2)  have  any  solution,  that  solution  is  x  =  2,  y  =  8. 
But  conversely,  x  =  2,  ?/  =  8  is  a  solution  of  (1),  (2),  inasmuch  as  the 
process  from  (1),  (2)  to  ^4),  (2)  is  reversible. 

Thus,  (3)  follows  from  (4),  and  then  (1)  from  (3),  (2). 

Note  1.     This  principle  of  substitution  is  a  consequence  of  the  several     364 
rules  of  equality,  §§  249,  253,  257,  and  of  the  general  rule  of  equality, 
If  a  =  6,  and  6  =  c,  then  a  =  c,  ^  261. 

Thus,  we  may  prove  our  right  to  make  the  preceding  substitution  as 
follows : 

liy  =  8,  then  ?/  +  2x  =  8  +  2x,  or2x  +  8  =  2x  +  7/,  §  249. 

And  if  2 X  +  8  =  2 X  +  2/,  and  2x  +  y  =  12,  then  2 x  +  8  =  12,  §  261. 

Note  2.     Of  course  this  principle  can  be  applied  only  when  we  have     365 
a  right  to  suppose  the  given  equations  to  be  simultaneous. 

Thus,  from  x  =  2  and  x  =  3  we  cannot  draw  the  absurd  conclusion 
2  =  3,  because  we  have  no  right  to  suppose  x  =  2,  x  =  3  simultaneous. 

TRANSFORMATION   THEOREMS 

In  view  of  what  has  just  been  said  we  may  regard  any     366 
correct  application  of  the  rules   of  reckoning  to  a  pair  of 
equations  as  a  legitimate  transformation  of  the  pair ;  and  if 
such  a  transformation  be  reversible,  we  may  conclude  that  it 
leaves  the  solutions  of  the  pair  unchanged. 

Hence  the  following  theorems,  which  hold  good  for  equa- 
tions in  any  number  of  unknown  letters. 

Theorem  1.      The  solutions   of  a  pair  of  equations   remain     367 
unchanged  when  the  transformations  o/  §§  338,  339  «r«  applied 
to  the  equations  separately. 


130  A   COLLEGE   ALGEBRA 

For  the  solutions  of  the  individual  equations  remain  unchanged  by 
such  transformations. 

Thus,  the  pair  of  equations  3  x  —  2  ?/  =  1  and  ?/  —  2  x  =  5 
has  the  same  solutions  as  3 x  —  2  y  =  1  and  y  =  5  +  2x. 

368  Theorem  2.      The  j^air  of  equations 

y  =  X,      f(x,y)=0 
has  the  same  solutio)is  as  the  pair 

y  =  X,      f(x,X)=0. 

Here  X  denotes  any  expression  in  x  alone  (or  a  constant),  /(x,  y)  any 
expression  in  x  and  y,  and/(x,  X)  the  result  of  substituting  X  for  y  in 
/(x,  y),  §280. 

The  theorem  is  merely  a  special  case  of  the  principle  of  substitution. 

Thus,  the  pair  of  equations  y  =  x  +  2  and  3  x  —  2  ?/  =  1 

has  the  same  solutions  as  ?/  =  x  +  2  and  3  x  —  2  (x  +  2)  =  1. 

369  Theorem  3.     The  pair  of  equations 

A  =  B,  C  =  D 

has  the  same  solutions  as  the  pair 

A  +  C  =  B  +  D,     C  =  D. 

For  A  =  B,  C  =  I)  has  the  same  solutions  as  A  +  C  =  B  +  C, 
C  =  D,  %  338,  and  A  +  C  =  B  +  C,  C  =  D  has  the  same  solutions  as 
A  +  C  =  B  +  n,  C  =  D,  ^  3(53. 

Thus,  the  pair  x  +  y  =  5  and  x  —  ?/  =  1 

has  the  same  solution  as   x  +  ?/  +  (x  —  ?/)  =  5  +  1  and  x  —  //  =  1, 
and  therefore  as  2x  =  (i  and  x  -  y  =  1. 

370  Corollary.  Before  applying  the  theorem  of  §  369  we  may, 
without  changing  their  solutions,  multiply  both  the  given 
equations  —  that  is,  both  members  of  each  equation  —  by  any 
constants  we  please,  except  0.     Hence 

If  k  and  1  denote  amj  constants  except  0,  the  pair  of  equations, 
A  =  B,  C  =  D 

has  the  same  solutions  as  the  pair 

kA  ±  IC  =  kB  ±  ID,     C  =  D. 


SIMULTANEOUS    SIMPLE    EQUATIONS  131 

Theorem  4.      When  A,  B,  and  C  are  integral,  the  pair  of     371 
equations  AB  =  0,  C  =  0 

has  the  same  solutions  as  the  tivo  pairs 

A  =  0,   C  =  0  and  B  =  0,   C  =  0. 

For  AB  —  0  has  the  same  solutions  as  the  two  equations  ^  =  0  and 
B  =  0  jointly,  §  341. 

Hence  the  solutions  of  the  pair  AB  =  0,  C  =  0  are  the  same  as  those 
of  the  pairs  ^  =  0,  (7  =  0  and  ^  =  0,  C  =  0  jointly. 

Thus,  the  solutions  of  xy  =  0  and  x  +  y  =  2 
are  that  of  x  =  0  and  x  +  y  =  2, 

together  with  that  of  y  =  0  and  x  +  y  =  2. 

Equivalent  systems.     Two  systems  of  simultaneous  equations     372 
are  said  to  be  equivalent  when  their  solutions  are  the  same. 

Thus,  the  pair  of  equations  x  +  2z/  =  5,  2x  +  2/  =  4is  equivalent  to  the 
pair  3x  +  ?/  =  5,  4x  +  32/  =  10,  both  pairs  having  the  same  solution  1,  2. 

Again,  the  pair  xy  =  0,  x  +  ?/  =  2  is  equivalent  to  the  two  pairs  x  =  0, 
X  +  y  =  2  and  y  =  0,  x  +  y  —  2. 

ELIMINATION.     SOLUTION  OF  A  PAIR  OF  SIMPLE  EQUATIONS 

Elimination.     To  eliminate  an  unknown  letter,  as  x,  from  a     373 
pair  of  equations  is  to  derive  from  this  pair  an  equation  in 
which  X  does  not  occur. 

We  proceed  to  explain  the  more  useful  methods  of  elimi- 
nating X  01  y  from  a  pair  of  simple  equations  in  x  and  y,  and 
of  deriving  the  solution  of  the  equations  from  the  result. 

Method  of  substitution.     This  method  is  based  on  the  theorem     374 
of  §  368. 

Example.     Solve  x  +  3  ?/  =  3,  (1) 

3x  +  5y  =  l.  (2) 

Solving  (1)  for  x  in  terms  of  ?/,  x  =  3  —  3  ?/.  (3) 

Substituting  3  -  3  ?/  for  x  in  (2),  3  (3  -  3  2/)  +  5  ?/  =  1.  (4) 

Solving  (4),  y  =  2.  (5) 

Substituting  2  for  2/  in  (3),  x  =  -  3.  (6) 


132  A   COLLEGE   ALGEBRA 

Hence  the  solution,  and  the  only  one,  of  (1),  (2)  is  x  =  —  S,  y  —  2. 

For,  by  §§  367,  3(j8,  the  following  pairs  of  equations  have  the  same 
3olution,  namely:  (1),  (2);  (3),  (2);  (3),  (4);  (3),  (5);  (6),  (5);  and  the 
solution  of  (6),  (5)  is  x  =  —  3,  ?/  =  2. 

The  same  conclusion  may  be  drawn  directly  from  §  362.  For  the 
process  from  (1),  (2)  to  (5),  (6)  is  reversible. 

Verijicatioji.      -3  +  3-2  =  3,     (1)  3  ( -  3)  +  5  •  2  =  1.     (2) 

Here  (4)  w^as  obtained  by  eliminating  x  by  substitution. 

To  eliminate  an  unknown  letter,  as  x,  from  a  pair  of  equations 
by  substitution,  obtain  an  expression  for  x  in  terms  of  the  other 
'    letter  (or  letters)  from  one  of  the  equations,  and  then  in  the  other 
equation  replace  x  by  this  expression. 

375  The  following  example  illustrates  a  sjDCcial  form  of  this 
method,  called  elimination  by  comparison. 

Example.     Solve                                        x  +  5?/  =  7,  (1) 

a;  +  6  2/  =  8.  (2) 
Solving  both  (1)  and  (2)  for  x  in  terms  of  ?/, 

x  =  7-5y,       (3)                               x  =  8-67/.  (4) 

Equating  these  two  expressions  for  x,    7  —  5  ?/  =  8  —  6  y.  (.5) 

Solving  (5)                                                              y=\.  (0) 

Substituting  1  for  y  in  (3),                                  x  =  2.  (7) 
Hence  the  solution  of  (1),  (2)  is  x  =  2,  ?/  =  1. 

376  Method  of  addition  or  subtraction.     Tliis  method  is  based  on 
the  theorem  of  §§  3G9,  o70. 


Example.     Solve 

2x-G?/  =  7, 
3x  +  4?/  =  4. 

(1) 
(2) 

Multiply  (1)  by  3, 

Gx-  18?/  =  21. 

(3) 

Multiply  (2)  by  2, 

Gx+    8v=    8. 

(4) 

Subtract  (4)  from  (3), 

-26y  =  13. 

(5) 

Whence, 

2/ =  -1/2. 

(6) 

Substitute  -  1/2  for  y  in  (1) 

2x 

-G(-l/2)  =  7. 

(7) 

Whence, 

x  =  2. 

(8j 

Hence  the  solution  of  (1), 

(2)  i&x^ 

=  2,  2/ =-1/2. 

SIMULTANEOUS    SIMPLE    EQUATIONS  133 

For  by  §§  367,  368,  370,  the  following  pairs  of  equations  have  the  same 
solution,  namely  :  (1),  (2) ;  (1),  (5) ;  (1),  (6) ;  (7),  (6) ;  (8),  (6) ;  and  the 
solution  of  (8),  (6)  is  x  =  2,  ?/  =  -  1/2. 

Verification.     2  •  2  -  6(- 1/2)  =  7,  (1)    3  •  2  +  4(- l/2)  =  4.      (2) 
Here  x  was  eliminated  by  subtraction. 

We  can  also  find  the  value  of  z  directly  from  (1),  (2)  by  eliminating 
y  by  addition.     Thus, 

Multiply  (1)  by  2,  4x-12y  =  14.  (9) 

Multiply  (2)  by  3,  9x  +  12y  =  12.  (10) 

Add  (9)  and  (10),  13 x  =26.  (11) 

Whence,  as  before,  x  =2.  (12) 

To  eliminate  an  nnknoivn  letter,  as  x,  from  a  pair  of  simple 

equatiotis  by  addition  or  subtraction,  multiply  the  equations  by 

nuinbei's  which  will  make  the  coeffiycients  of  x  in  the  resulting 

equatio7is  equal  numerically.      Then  subtract  or  add  according 

as  these  coefficients  have  like  or  xinlike  signs. 

Exceptional  cases.  Let  A  =i),  B  =  0  denote  a  pair  of  simple  377 
equations  in  x  and  y.  Tlie  preceding  sections,  §§  374,  376, 
show  that  this  pair  ^  =  0,  J5  =  0  has  one  solution  and  but  one, 
unless  the  expressions  A  and  B  are  such  that  in  eliminating  x 
we  shall  at  the  same  time  eliminate  y.  This  can  occur  in  the 
following  cases  only. 

1.  If  the  expressions  .1  and  B  are  such  that  A  =  kB,  where  k 
denotes  a  constant,  we  say  that  the  equations  ^1  =  0  and  B  =  ^ 
are  not  independent. 

Evidently  if  J.  =  kB,  every  solution  of  £  =  0  is  a  solution 
of  J.  =  0,  and  vice  versa,  so  that  the  pair  ^  =  0,  i?  =  0  has 
infinitely  many  solutions. 

Thus,  let  ^  =  2 X  +  6 ?y  -  10  =  0  (1),  and  B  =  X  +  3 ?/  -  5  =  0  (2). 

Here  ^  =  2  /J,  so  that  yl  =  0  and  i?  =  0  are  not  independent.  Observe 
that  if  to  eliminate  x  we  multiply  (2)  by  2  and  subtract  the  result  from 
(1)  we  at  the  same  time  eliminate  y. 

2.  If  A  and  B  are  such  that  A  =  kB  -\-  I,  where  k  and  I 
denote  constants,  I  not  0,  we  say  that  the  equations  ^  =  0 
and  .6  =  0  are  not  consistent. 


134  A   COLLEGE   ALGEBRA 

In  this  case  the  pair  A  =  0,  B  =  0  has  no  solution ;  for  any 
values  of  x,  y  that  make  B  =  0  will  make  A  =  I,  not  ^  =  0. 

Thus,  \QtA  =  2x  +  Qy  -9  =  0  (3),  andB  =  a;  +  32/-5  =  0  (4). 
Here  J.  =  2  B  +  1,  so  that  A  =  0  and  B  =  0  are  not  consistent.     If  we 
eliminate  x  from  (3),  (4)  we  shall  at  the  same  time  eliminate  y. 

378         Formulas  for  the  solution.     We  may  reduce  any  given  pair  of 
simple  equations  in  x,  y  to  the  form 

ax  -{-  by  =  c,  (1)  a'x  +  b'y  =  e',  (2) 

where  a,  b,  c,  a',  b',  c'  denote  known  numbers  or  expressions. 

By  §  377,  the  pair  (1),  (2)  has  one  solution,  and  but  one, 
unless  a  constant  k  can  be  found  such  that  a'  =  ka  and  b'  =  kb, 
and  therefore  ab'  —  a'b  =k(ab  —  ab)=  0. 

To  obtain  this  solution,  eliminate  y  and  x  independently  by 
the  method  of  subtraction,  §  376.     The  results  are 

(ab'  —  a'b)  x  =  b'c  ~  be',   (3)       (ab'  —  a'b)  y  =  ac'  —  a'c.   (4) 

Therefore,  if  ab'  —  a'b  ^  0,  the  solution  of  (1),  (2)  is 

b'c  —  be'  ae'  —  a'e 


ab'  —  a'b  ab'  —  a'b 

These  formulas  are  more  easily  remembered  if  written 
^       _       y       ^      -1 

be'  —  b'c       ca'  —  c'a       ab'  —  a'b 


(5) 


(6) 


Did  we  not  know  in  advance  that  the  pair  (1),  (2)  has  a  solution  when 
ab'  —  a'b  ^  0,  the  argument  here  given  would  only  prove  that  if  the  pair 
(1),  (2)  has  any  solution,  it  is  (5). 

EXERCISE  VII 

Solve  the  following  pairs  of  equations  for  x  and  y. 

j'x  +  y  =  62,  j'6x-52/  =  2r),  j-45x-137/ =161, 

^'    |x-2/=12.  ■    l4x-3y  =  10.  '    1  18x  +  ll  y  =  32. 

x-3  =  7-x,  j'12x  =  9- lOz/,  ('22/-3x  =  0, 

8x-32/-61  =  0.       ■    t82/  =  7-9x.  '    t5x-32/-2  =  0. 


SIMULTANEOUS    SIMPLE    EQUATIONS  135 

fx/.S  +  by  =  S\,  (2{2x  +  Sy)  =  S{2x-3y)  +  10, 

L5x  +  32/  =  1.65.  ■     l4x-3?/  =  4(6y- 2x)  +  3. 

r{x  +  2){y  +  l)  =  (z-5){y-l),  (  ax  +  by  =  a'^  +  2a  +  b^, 

L X (4 +  ?/)  =  - 2/(8 -X).  ■     lbx  +  ay  =  a^  +  2b  +  b'^. 

Cax  +  by::^c,  f  {a-b)x  +  (a  +  b)y  =  2(a^-b^), 

\(a  +  b)x+(a~b)y  =  2{a:^  +  b'^). 

^'  r^-y  _  x  +  2?/-5  _  y-S  _  y  +  2x-5 

13.    "I"  "  14.    J      4  6         ~     4  6        ' 

5x-2y  +  6  =  0. 


'  ax  +  by  —  c, 

\px  =  qy. 

fx  +  yy- 

3              2 

X 

x      x  +  y 
^2"^      9 

7 

{^1  =  1 
a      b      c 

'   x      y  _\ 
a       b'      c' 

15.    ^  16. 


-  +  ^  =  l+x, 
a      b 

X      V 
,  V  +  -  =  1  +  2/- 
\^b      a 

17.  Show  that  the  following  equations  are  inconsistent. 

llx  -23  2/  =  10,  6z-10y  =  15. 

18.  In  Ex.  15  assign  values  to  a,  6,  c,  a',  6',  c'  for  which  the  equations 

are  (1)  not  consistent,  (2)  not  independent. 

PAIRS    OF   EQUATIONS    NOT   OF   THE   FIRST   DEGREE  WHOSE 

SOLUTIONS    CAN    BE   FOUND   BY   SOLVING   PAIRS 

OF   SIMPLE   EQUATIONS 

A  pair  of  equations  which  are  not  of  the  first  degree  with     379 
respect  to  x  and  y  may  yet  be  of  the  first  degree  with  respect 
to  a  certain  pair  of  functions  of  x  and  y.     We  can  then  solve 
the  equations  for  this  pair  of  functions,  and  from  the  result  it 
is  ofiten  possible  to  derive  the  values  of  x  and  y  themselves. 

1    ,      r.  ,       2        5       ,        9      10      , 

Example  1.     Solve  -H =  1,       -H =  5. 

X      3y  X       y 

Both  equations  are  of  the  first  degree  with  respect  to  1  /x  and  1/y. 
Solving  for  1/x  and  1/y,  we  find  l/x  =  1/3,  1/2/  =  1/5.     Hence 
x  =  3,  2/ =  5. 


136  A   COLLEGE   ALGEBRA 

Example  2.     Solve  3x  +  -  =  6,  7x =  1. 

Solving  for  x  and  y/x,  we  find  x  =  1,  rj/x  =  3.     Hence  x  =  1,  y  =  3. 

380  Given  a  pair  of  equations  reducible  to  the  form  AB  =^  0, 
A'B'  =  0,  where  A,  B,  A',  B'  denote  integral  expressions  of  the 
first  degree  in  x  and  y.  It  follows  from  the  theorem  of  §  371 
that  all  the  solutions  of  this  pair  can  be  obtained  by  solving 
the  four  pairs  of  simple  equations  A  =  0,  ^'  =  0;  A  =0, 
B'  =  0;  B  =  0,  A'  =  0;  B  =  0,   B' =  0. 

Example.     Solve                             x^  — 2xy  =  0,  (1) 

(x  +  y-l){2x+y-S)=0.  (2) 
This  pair  is  equivalent  to  the  four  pairs 

X  =  0,           X  +  y  -  1  =  0,  (3) 

X  =  0,         2  X  +  ?/  -  3  =  0,  (4) 

X  -  2  2/  =  0,           X  +  7/  -  1  =  0,  (5) 

x-22/  =  0,         2x  +  y  -5  =  0.  (6) 

Solving  these  four  pairs  (3),  (4),  (5),  (0)  we  obtain  the  four  solutions 

of  (1),  (2),  namely:  x,  y  =  0,  1  ;  0,  3 ;  2/3,  1/3;  G/5,  3/5. 

381  And,  in  general,  if  ABC  ■  ■  ■  and  A'B'C"  ■  ■  ■  denote  products  of 
m  and  n  integral  factors  of  the  first  degree  in  x  and  y,  all  the 
solutions  of  the  pair  of  equations  ABC  •  •  •  =  0,  A'B'C  •  •  •  =  0 
can  be  found  by  solving  the  mn  pairs  of  simple  equations 
obtained  by  combining  each  factor  of  the  first  product  equated 
to  0  with  each  factor  of  the  second  likewise  equated  to  0. 

If  all  these  pairs  of  simple  equations  are  both  independent 
and  consistent,  we  thus  obtain  mn  solutions,  that  is,  the  number 
of  solutions  of  the  given  equations  is  the  prod^ict  of  their  degrees. 

EXERCISE  Vra 

Solve  the  following  pairs  of  equations. 

2x      3?/  X       y 

2.    10x  +  5  =  5,       15x  +  — =  8. 

y  y 


SIMULTANEOUS    SIMPLE    EQUATIONS 


131 


y2(3-y)   ^  3 
X  X  2' 


X  X 

+  2)-- 


+  1. 
=  0. 


4.  12/ =  0,       (a;  +  2  2/-  l){3x 

5.  xy  -  y  =  0,       Sx  -8y  +  5  =  0. 

6.  X  (cc  -  ?/)  (X  +  ?/)  =  0,       X  +  2  2/  -  5  =  0. 

7.  (X  -  1)  (y  -  2)  =  0,       (X  -  2)  (2/  -  3)  =  0. 

8.  2/2  ={x- 1)2,       2x  +  32/-7  =  0. 

9.  (2x  +  2/)2=(x-3  2/  +  5)2,       (x  +  2/)2  =  1. 

10.  (x-52/  +  8)(x  +  32/  +  5)  =  0,    (2x  +  2/ +  5){5x  +  2?/ -  14)  =  0. 


GRAPHS   OF   SIMPLE  EQUATIONS    IN   TWO   VARIABLES 

Graph  of  a  pair  of  values  of  x  and  y.  It  is  convenient  to 
represent  /^a/?-*'  of  values  of  two  variables,  as  x  and  y,  by 
points  in  a  plane. 

In  the  plane  select  as  axes  of  reference  two  fixed  straight 
lines,  X^OX  and  Y'OY,  which  meet  at  right  angles  at  the  point 
0,  called  the  origin ;  and  choose 
some  convenient  unit  for  measuring 
lengths. 

Then  if  the  given  pair  of  val- 
ues be  a;  =  a,  y  =  h,  proceed  as 
follows  : 

On  X'OX  and  to  the  right  or  left 
of  O,  according  as  a  is  positive  or 
negative,  measure  off  a  segment,  0.4, 
whose  length  is  \a\,  the  numerical 
value  of  a. 

Similarly  on  Y'OY  and  above  or  below  0,  according  as  h  is 
positive  or  negative,  measure  off  a  segment,  OB,  whose  length 
is|/.|. 

Then  through  A  and  B  draw  parallels  to  Y'OY  and  X'OX 
respectively.     We  take  P,  the  point  in  which  these  parallels 


sY 


382 


138 


A    COLLEGE    ALGEBRA 


intersect,  as  the  point-picture,  or  graj^li^  of  the  pair  of  values 
X  ^^  a,  y  =  h. 

It  is  convenient  to  represent  both  the  value-pair  x  =  a,  y  =  b 
and  its  graph  P  by  the  symbol  (a,  b). 

We  call  the  number  a,  or  one  of  the  equal  line  segments 
OA  or  BP,  the  abscissa  of  P ;  and  b,  or  one  of  the  equal  seg- 
ments OB  or  AP,  the  ordinate  of  P.  And  we  call  the  abscissa 
and  ordinate  together  the  codrdi?iates  of  P. 

We  also  call  X'OX  the  x-axis  or  the  axis  of  abscissas,  and 
Y'OY  the  y-axis  or  the  axis  of  ordi^iates. 

Observe  that  this  method  brings  the  value-pairs  of  a;,  y  into  one-to-one 
correspondence,  §  2,  with  the  points  of  the  plane ;  that  is,  for  each  value- 
pair  (a,  b)  there  is  one  point  P,  and  reciprocally  for  each  point  P  there 
is  one  value-pair  (a,  6)  found  by  measuring  the  distances  of  P  from  Y'OY 

and  X'OX  respectively,  and 
giving  them  their  appro- 
priate signs. 

In  particular,  the  graph 
of  (0,  0)  is  the  origin,  that 
of  (a,  0)  is  a  point  on  the 
z-axis,  and  that  of  (0,  h)  is 
a  point  on  the  y-axis. 

Example.  Plot  the  value- 
pairs  (4, 4),  (-3,  3),  (-4,0), 
(-  5,  -  4),  (3,  -  2). 

Carrying  out. the  construc- 
tion just  described  for  each 
value-pair  in  turn,  we  obtain 
their  graphs  as  indicated  in 
the  accompanying  figure. 
Notice  particularly  how  the  position  of  the  graph  depends  on  the  signs 
of  the  coordinates. 

383  The  graph  of  an  equati^  in  x  and  y.  If,  as  is  commonly  the 
case,  a  given  equation  m  x  and  y  has  infinitely  many  real 
solutions,  there  will  usually  be  a  definite  curve  which  con- 
tains the  graphs  of  all  these  solutions  and  no  other  points. 
We  call  this  curve  the  (jraph  of  the  equation. 


r 

(-3,3) 

1 



(4,4) 

• 

1 
1 

i 

1 
.,  ^  (-4,0)  1 

1 

'      •       '       ' 

0 

•  ■— T ' — :Y 

(3,-2) 

(-0 

-4) 

SIMULTANEOUS    SIMPLE    EQUATIONS 


139 


But  the  graph  of  an  equation  may  consist  of  more  than  one  curve. 
Observe  that  we  here  include  straight  lines  among  curves. 

Theorem.     The  graph  of  every  simple  equation  in  one  or  both 
of  the  letters  x  and  y  is  a  straight  line. 

On  this  account  simple  equations  are  often  called  linear  equations. 

The  student  may  readily  convince  himself  of  the  truth  of 
the  theorem  by  selecting  some  particular 
equation  and  "  plotting  "  a  number  of  its 
solutions. 


(-1 


X- 


6) 


(0,4) 


^IV) 


(2^0) 


(3,-2) 


Thus,  take  the  equation  ?/  =  —  2  x  +  4. 

When    x=:0,     1,     2,         3,     ••• 
we  have     ?/  =  4,     2,     0,     -  2,     •  •  • 

And  plotting  these  value-pairs  (0,  4),  (1,  2), 
(2,  0),  (3,  —  2)  •  •  •  as  in  the  accompanying 
figure,  we  find  that  their  graphs  all  lie  in  the 
same  straight  line. 

We  Ta?iY  prove  the  theorem  as  follows  : 
1.    When  the  equation  has  the  form 
X  =  a,  or  y  =  b. 

Example.     Find  the  graph  of  x  =  2. 

This  equation  is  satisfied  by  the  value  2  of  x  and  every  value  of  y,  §  356. 
Hence  the  graph  is  a  parallel  to  the  2/-axis  at 
the  distance  2  to  its  right.  For  this  line  con- 
tains all  points  whose  abscissas  are  2,  and  such 
points  only. 

And  so,  in  general,  the  graph  oi  x  =  a 
is  a  parallel  to  the  ^/-axis  at  the  distance 
\a\to  the  right  or  left  according  as  a  is 
positive  or  negative;  and  the  graph  of 
2/  =  i  is  a  pa  lei  to  the  x-axis  at  the 
distance  |^>|   above  or  below  according 


A'^ 


(x  =  2) 


as  b  is  positive  or  negative. 

In  particular,  the  graph  of  y 
ic  =  0  is  the  y-axis. 


0  is  the  X-axis,  and  that  of 


384 


140 


A   COLLEGE   ALGEBRA 


385 


2,    When  the  equation  has  the  form  y  =  mx. 
Example.     Find  the  graph  of  y  =  2  x. 

The  graph  is  the  right  line  which  passes  through  the  origin  (0,  0)  a,nd 
the  point  (1,  2) ;  for  this  line  contains  every  point  whose  ordinate  is  twice 
its  abscissa,  and  such  points  only. 

And  so,  in  general,  the  graph  of 
y  =  mx  is  the  right  line  which  passes 
hrough  the  origin  and  the  point  (1,  m). 

3.    When  the  equation  has  the  form 
y  =  mx  +  c. 

Example.     Find  the  graph  of  y  =  2  x  +  3. 
Evidently  we  shall  obtain  the  graph  of  this 
equation  if  we  increase  the  ordinate  of  every 
point  of  the  graph  of  y  =  2x  by  3.     But  that  comes  to  the  same  thing 
as  shifting  the  line  y  =  2x  upward  parallel  to 
itself  until  its  point  of  intersection   with  the 
{/-axis  is  3  units  above  the  origin. 

And  so,  in  general,  the  graph  of 
y  =  mx  +  c  is  a  right  line  parallel  to 
the  graph  of  y  =  mx  and  meeting  the 
y-axis  at  the  distance  \c\  from  the  origin, 
above  or  below,  according  as  c  is  posi- 
tive or  negative. 

To  find  this  line.     As  any  two  of  its 
points  suffice  to  determine  a  right  line, 
we  may  find  the  graph  of  any  equation,  ax  -\-  by  +  c  =z  0,  as 
in  the  following  example. 

Example.     Plot  the  graph  of3a;  +  y  —  6  =  0. 

First,  when  y  =  0,  then  x  =  2.     Second,  when  x  =  0,  then  y  =  6. 

Hence  we  have  only  to  plot  the  points  (2,  0)  and  (0,  6),  that  is,  the  points 
where  the  line  will  meet  the  axes,  and  draw  the  line  which  these  points 
deternnne  (see  figure  in  §  380). 

This  method  fails  when  the  equation  has  one  of  the  forms  x  =  a, 
y  =  b,  y  -  mx.  We  then  find  the  line  by  the  methods  explained  in 
§  384,  1  and  2. 


SIMULTANEOUS   SIMPLE   EQUATIONS 


141 


\ 

r 

\ 

<5' 

\ 

V- 

\^   Xe/^^'-' 

^J>^S,4) 

M\^,3)  \       ^^;> 

V' 

\          :?'^5, 1) 

,'' 

\ 

Example.     Find  the  graphs  of  the  equations 

{4x  +  2/-7)(3x  +  2y-17)  =  0, 
(X  -  2  y  +  6)  (2  X  -  3  2/  -  7)  =  0, 
and  the  graphs  of  their  solutions. 


387 


Graph  of  the  solution  of  a  pair  of  simultaneous  simple  equations.     386 

This  is  the  point  of  intersection  of  the  two  lines  which  are  the 

graphs  of  the  equations  themselves  ; 

for  this  point,  and  this  point  only, 

is  the  graph  of  a  solution  of  both 

equations. 

Thus,  the  solution  of2x-3y  +  7  =  0 
(l),and3x  +  2/-6  =  0(2)isx  =  l,  y  =  3. 
And,  as  the  figure  shows,  the  graphs  of  (1) 
and  (2)  intersect  at  the  point  (1,  3). 

When  the  given  equations  are  not 
consistent,  §  377,  2,  their  graphs  are  A' 
lines  which  have  no  point  in  common, 
thatis,j9araZZenines;  when  theequa- 
tions  are  not  independent,  §  377,  1, 
their  graphs  are  lines  which  have  all  their  points  in  commonj, 

that  is,  coincident  lines. 

Thus,  the  equations  ?/  =  2  x, 
y  =  2  X  +  3  are  not  consistent, 
and  the  graphs  of  these  equa- 
tions, §  384,  3,  are  parallel  lines. 

Again,  the  equations?/  =  2x, 
3  ?/  =  6  X,  which  are  not  inde- 
pendent, have  the  same  graph. 

The  graph  of  an  equa-  388 
tion  of  the  form  AB  =  0 
consists  of  the  graphs  of 
^  =  0  and  B  =  0  jointly ; 
for  the  solutions  oi  AB  =  0 
are  those  of  J  =  0  and  B  =  0 
jointly,  §§  341,  346. 


(1) 
(2) 


142 


A   COLLEGE   ALGEBRA 


The  graph  of  (1)  consists  of  the  lines  PQ  and  ES  which  are  the  graphs 
of4x  +  y  —  7=0  and  3x4-2?/  —  17=0  respectively. 

The  graph  of  (2)  consists  of  the  lines  PS  and  QR  which  are  the  graphs 
ofx  —  2?/  +  5  =  0  and  2x  —  3?/  —  7  =  0  respectively. 

The  points  P,  Q,  E,  S  in  which  the  pair  PQ,  RS  meets  the  pair 
PS,  QR  are  the  graphs  of  the  solutions  of  (1),  (2),  namely,  (1,  3),  (2,  -  1), 
(5,  1),  (3,  4). 

389  Graph  of  an  equation  of  higher  degree  in  x  and  y.  We  find  a 
number  of  the  solutions  of  the  equation,  plot  these  solutions, 
and  then  with  a  free  hand  draw  a  curve  which  will  pass 
through  all  the  points  thus  found.  By  taking  the  solutions 
"  near "  enough  together,  we  can  in  this  way  obtain  a  curve 
which  differs  from  the  true  graph 
as  little  as  we  please. 

In  work  of  this  kind  it  is  con- 
venient to  use  paper  ruled  into 
small  squares,  as  in  the  accom- 
panying figure. 

Example.  Find  the  graph  of  the 
equation  y  =  x-. 

When   x  =  0,     1,      2,  8,       4,  ■  •  • 

we  have    y  =  0,     1,      4,  9,     16,  •  •  • 

And  when      -  1,  -  2,  -  3,  -  4,  ••  • 

we  have                  1,       4,  9,     10,  •  •  • 

Taking  the  side  of  a  square  as  the 
unit  of  length,  plot  the  corresponding  points  (0,  0),  (1,  1),  (2,  4)  ■  •  •(—  1, 1), 
(—  2,  4)-  •  •.  A  few  of  them  suffice  to  indicate  the  general  character  of 
the  graph,  the  curve  in  the  figure,  except  between  x  =  —  1  and  x  =  +  1. 

It  lies  wholly  above  the  x-axis,  extending  upward  indefinitely  ;  and  it 
lies  symmetrically  with  respect  to  the  y-axis,  the  same  value  of  y  corre- 
sponding to  X  =  a  and  x  =  —  a. 

When  x  =  ±l/2,      ±1/4,... 

we  have  2/ =      1/4,       1/16,  ••• 

and  plotting  one  or  two  of  the  corresponding  points,  we  find  that  the 
graph  touches  the  x-axis. 


' 

i 

\ 

\ 

\ 

\ 

1 

\ 

1 

\ 

\ 

/ 

, 

k 

y 

0 

SIMULTANEOUS    SIMPLE    EQUATIONS  143 

EXERCISE  IX 

1.  Plot  the  following  pairs  of  values  of  x  and  y. 

(0,  0),  (5,  0),  (0,  -  7),  (6,  2),  (-  7,  -  1),  (-  4,  3),  (5,  -  9). 

2.  Find  the  graphs  of  the  following  equations. 

x  =  0,       y  =  0,       2  2/  +  7  =  0,       3(/  +  x  =  0,      x  +  2/  +  5  =  0, 
7x4-3?/- 18  =  0,       3x-4y  =  24. 

3.  Find  the  graphs  of  the  following. 

X2/  =  0,     (X  4-  y  -  3)  (X  -  2  2/)  =  0,     x2  -  1  =  0,    x2  =  4  y"^,     x"^ -\- y"^  =  0. 

4.  Find  the  solutions  of  the  following  pairs  of  equations  by  the  graphi- 
cal method  and  verify  the  results  algebraically. 

rx  +  2/-3=:0,  r32/  +  2x  +  19  =  0, 

^'\      x-2?/  =  0.  ^  '    t22/-3x  +  4  =  0. 

5.  Do  the  same  with  each  of  the  following  pairs. 

r'(x-4y  +  6)(x  +  32/  +  6)  =  0,  r-(y_x-2)x  =  0, 

t(3x  +  22/-10)(2x-2/  +  5)  =  0.  I  (y  -  x  +  2)2/ =  0. 

6.  Find  the  graphs  of  the  following  two  equations. 

2/  =-(x+  1)2,  2/  =  x\ 


SYSTEMS   OF  SIMPLE   EQUATIONS  WHICH  INVOLVE   MORE 
THAN   TWO   UNKNOWN   LETTERS 

Method  of  solving  a  system  of  n  simple  equations  in  n  unknown     390 
letters.     A  pair  of  equations  in  three  unknown  letters  will 
ordinarily  have  infinitely  many  solutions. 

Thus,  the  pair  x  =  2  2;,  y  =  z  +  \  has  infinitely  many  solutions ;  for 
both  equations  are  satisfied  if  we  assign  any  value  whatsoever,  as  6,  to  z 
and  the  values  2  h  and  &  +  1  to  x  and  2/- 

But  a  system  of  three  simple  equations  in  three  unknown     391 
letters  ordinarily  has  one,  and  but  one,  solution,  which  may 
be  obtained  as  in  the  following  example. 
Example.     Solve  the  system  of  equations 

3x-22/  +  4z  =  13,  (1) 

2  X  +  5  ?/  -  3  z  =  -  9,  (2) 

6x  +  3y  +  23  =  7.  (3) 


144  A    COLLEGE   ALGEBRA 

Elimiuate  z  between  two  pairs  of  these  equations,  thus : 
Multiply  (1)  by  3,         9  x  -    6  y  +  12  2  =  39  (4) 

Multiply  (2)  by  4,         8x  +  20?/-  122  =-36  (5) 

Add  ITx+Hy  =3  (6) 

Again,  (1)  is  3x-2y  +  42=13  (7) 

Multiply  (3)  by  2,  12x  +  6y  +  42  =  14  (8) 

Subtract  (7)  from  (8),       9x  +  8y  =1  (9) 

Eliminate  y  between  the  resulting  equations  (6),  (9),  thus: 
Multiply  (6)  by  4,  68  x  +  56  ?/  =  12  (10) 

Multiply  (9)  by  7,  63x  +  56i/=    7  (11) 

Subtract  (11)  from  (10),  5x  =5  (12) 

Hence  x  =  1. 

Substituting  x  =  1  in  (9),  we  find  y  —  —\. 

Substituting  x  =  1,  y  =  —  1  in  (1),  we  find  2  =  2. 

Therefore,  §  362,  if  (1),  (2),  (3)  has  any  solution,  it  is  x  =  1,  y  =  -  1, 
2  =  2.  But  the  process  by  which  we  have  dei-ived  x  =  l,y  =  — 1,2  =  2 
from  (1),  (2),  (3)  is  reversible.  In  fact,  it  may  readily  be  traced  back- 
ward step  by  step.  Hence  x  =  1,  y  =  -  1,  2  =  2  is  the  solution  of  (1), 
(2),  (3). 

We  may  also  prove  as  follows  that  x  =  1,  y  =  —  1,  2  =  2  is  the  solution 
of  (1),  (2),  (3). 

It  is  evident  by  §  368  that  x  =  1,  y  =  -  1,  2  =  2  is  the  solution  of  (12), 
(9),  (1).  We  therefore  have  only  to  prove  that  the  system  (12),  (9),  (1) 
has  the  same  solution  as  the  given  system  (1),  (2),  (3). 

Let  us  represent  the  equations  (1),  (2),  (3),  with  the  known  terms  trans- 
posed to  the  first  members,  thus  : 

^  =  0,     (1)  B  =  0,     (2)  C  =  0.     (3) 

It  will  then  follow  from  the  manner  in  which  (9)  and  (12)  were  derived, 
that  we  may  express  the  equations  (1),  (9),  (12)  thus: 

^=0,    (1)       -^  +  2C  =  0,     (9)       19^ +  16B-14C  =  0.     (12) 

Evidently  any  set  of  values  of  x,  y,  z  that  makes  A=0,  B  =  0,  C  =  0 
will  make^^O,  -A  +2C  =  0,  19  A  +  16B-UC  =  0. 

Conversely,  when  A=0  and  —A  +2C  =  0,  then  C  =  0;  and  when 
also  19  ^  +  1(5  B-14C  =  0,  then  B  =  0. 

Hence  the  system  (1),  (2),  (3)  has  the  same  solution  as  the  system  (1), 
<9),  (12),  namely,  x  =  1,  y  =  -  1,  2  =  2. 


SIMULTANEOUS    SIMPLE    EQUATIONS  145 

In  the  case  just  considered,  from  the  given  system  of  three    392 
equations  in  the  three  unknown  letters,  x,  y,  z,  we  derived  a 
system  of  tivo  equations  in  two  letters,  x,  y,  and  then  from 
this  system,  a  single  equation  in  one  letter,  x. 

And,  in  general,  if  we  start  with  a  system  of  n  simple  equa- 
tions in  n  unknown  letters  and  take  n  —  1  oi  these  steps,  wg 
shall  arrive  at  a  single  equation  in  one  of  the  letters,  as  x,  of 
the  form  ax  —  h  =  0. 

Then,  ujiless  a  =  0,  the  system  has  one,  and  hut  one,  solution, 
in  which  the  value  of  x  is,  b /a  and  the  values  of  the  other 
unknown  letters  may  be  found  by  successive  siibstitutions  in 
the  equations  obtained  in  the  process.  This  may  always  be 
proved  as  in  the  example. 

On  the  other  hand,  if  a  =  0  the  system  ordinarily  has  infi- 
nitely many  solutions  when  b  =  0,  and  no  solution  when  b  ^  0. 
This  will  be  proved  in  §  394. 

A  much  less  laborious  method  of  solving  a  system  of  simple     393 
equations  is  given  in  the  chapter  on  determinants.     In  certain 
cases  labor  may  be  saved  by  special  devices. 

Example.     Solve  x  + 1/  +  2  =    8,  (1) 

X  +  2/  +  ?t  =  12,  (2) 

X  +  2  +  tt  =  14,  (3) 

7J  +  z  +  u  =  14,  (4) 

Add  (1),  (2),  (3),  (4),  3x  +  3?/  +  32  +  3u  =  48. 

Hence  x  -\-  y  +  z  +  u  =  16.  (5) 

And  subtracting  each  of  the  equations  (4),  (3),  (2),  (1)  in  turn  from  (5), 
we  obtain  x  =  2,  y  =  2,  2  =  4,  m  =  8. 

Exceptional  cases.     Let  A=Q,  B  =  0,  C  =  0  denote  a  system     394 
of  simple  equations  in  x,  y,  z,  and,  as  in  §  392,  let  ax  —  b  —  ^ 
denote  the  equation  obtained  by  eliminating  y  and  z. 

1.  If  a  =  0  and  J  =  0,  it  will  be  found  that  one  of  the  func- 
tions A,B,C  may  be  expressed  in  terms  of  the  other  two,  thus  : 
A^kB  -\-  IC,  where  k  and  I  denote  constants.  We  then  say 
that  the  equations  A  =  d,  B  —  0,  C  —  0  are  not  Independent. 


146  A    COLLEGE    ALGEBRA 

From  the  identity  A  =  IB  +  IC  it  follows  that  every  solu- 
tion oi  B  =  0  and  C  =  0  is  a  solution  of  ^  =  0.  Hence  if 
^  =  0  and  C  —  0  are  consistent,  §  377,  2,  the  three  equations 
A  =  0,  B  =  0,  C  =  0  will  have  infinitely  many  solutions. 

Thus,  consider  the  system  of  equations 

^  =  3a;-2y  +  42-13  =  0,  (1) 

£=:2x  +  5y-3z+9  =  0,  (2) 

C=7x  +  8  7/-2z+5  =  0.  (3) 

Eliminating  z  between  (1)  and  (2), 

,     3^ +4B=17x  +  14  2/-3  =  0.  (4) 

Eliminating  z  between  (1)  and  (3), 

^  +  2Cee17x  + 142/-3  =  0.  (5) 

Eliminating  y  between  (4)  and  (5), 

24  +  4B-2C=       0-x        -0  =  0.  (6) 

Here  the  final  equation  ax  —  6  =  0  has  the  form  0  •  x  —  0  =  0,  and  in 
deriving  it,  we  find  that  the  expressions  A,  B,  C  are  connected  by  the 
identity  2  A  +  i  B  -  2  C ~0,  or  C  =  A  +  2  B. 

And,  in  fact,  we  see  on  examining  (1),  (2),  (3)  that  C  may  be  obtained 
by  multiplying  5  by  2  and  adding  the  result  to  A. 

Hence  the  system  (1),  (2),  (3)  has  infinitely  many  solutions. 

2.  If  a  =  0  and  b  ^  0,  it  will  be  found  that  one  of  the  func- 
tions A,  B,  C  may  be  expressed  in  terms  of  the  other  two,  thus  : 

A  =  IcB  +  IC  +  m, 

where  k,  I,  vi  denote  constants,  m  not  0.    We  then  say  that  the 
equations  A  =  0,  B  =  0,  C  =  0  are  7iot  consistent. 

t'rom  the  identity  A  =  kB  +  IC  +  m  it  follows  that  ^  =  0, 
B  =  Q,  C  =  0  have  no  solution.  For  any  values  of  x,  y,  z  that 
make  B  =  0  and  C  =  0  will  make  A  =  m,  not  A  =  0. 

Thus,  consider  the  system  of  equations 

A  =  3x-2y  +  iz-lS  =  0,  (1) 

B  =  2x  +  5y  -Sz+    0  =  0,  (2) 

C  =  7x-l-8y-2z+    6  =  0.  (3) 


SIMULTANEOUS   SIMPLE   EQUATIONS  147 

Eliminating  z  and  y  as  above,  we  obtain 

2^+4B-2C  =  0x-2  =  0. 

Hence  the  final  equation  ax  —  6  =  0  has  the  form  0  •  x  —  2  =  0,  and 
A,  B,  C  are  connected  by  the  identity  C  =  ^  +  2^+L  And,  in  fact, 
on  examining  (1),  (2),  (3)  we  find  this  to  be  the  case. 

Hence  the  system  (1),  (2),  (3)  has  no  solution. 

Systems  of  simple  equations  in  general.     From  the  preceding     395 
discussion  we  may  draw  the  conclusion : 

Ordinarily  a  system  of  m  siynple  equations  in  n  unknown 
letters  has  one  solutio7i  tvJien  m  =  n,  infinitely  many  solutions 
when  m  <  n,  no  solution  ivhen  m  >  n. 

JVhenever  exceptions  to  this  rule  occur,  two  or  more  of  the  equa- 
tions are  connected  hy  identical  7'elations  of  the  kind  described  in 
§§  377,  394. 

In  particular,  a  system  of  three  simple  equations,  ^  =  0, 
B  =  0,  C  =  0,  in  tivo  unknown  letters,  oc,  ?/,  has  a  solution 
when,  and  only  when,  A,  B,C  are  connected  by  an  identity  of 
the  form  A  =  kB  -{-  IC  and  B—0,  C  =  0  are  consistent. 

Thus,  the  system  x  -  y  =  1  (1),  x  +  y  =  1  (2),  3x  -  y  =  10  (3)  has  no 
solution  ;  for  the  solution  of  (1),  (2),  namely,  x  =  4,  y  =  3,  does  not 
satisfy  (3). 

On  the  other  hand,  the  system  x-?/  =  l(l),x  +  y  =  7  (2),  3  x  -  y  =  9  (4) 
has  a  solution  ;  for  (4)  is  satisfied  by  x  =  4,  y  —  S.  But  observe  that 
3x  -  y  -  9  =  2  (X  -  2/  -  1)  +  (X  +  2/  -  7). 

Let  the  student  draw  the  graphs  of  (1),  (2),  (4).  He  will  find  that  they 
meet  in  a  common  point. 

EXERCISE   X 

Solve  the  following  systems  of  equations. 
(x  +  y=n, 
1.    J  2/  +  z  =  13, 
[z  +x  =  12. 

fx-f22/-3z  =  3, 
3.   J  3x-52/  +  7z  =  19, 
Ux-Sw-  llz=:-  13. 


x+y+z= 

1, 

x  +  2y  +  Sz 

=  4, 

x  +  3?/  +  72 

=  13. 

5x-22/  =  - 

-33, 

X  +  2/  -  7  z  - 

=  13, 

x+3y=- 

10. 

148 


A   COLLEGE   ALGEBRA 


X  +  2y  —  4z 
2  X  -  3  y  =  0, 
y  -  4  z  =  0. 


11, 


(3x-5  =  2(x-2), 
(x+l)(y-l)-(x  +  2)(y-2)  +  5, 
2x  +  Sy  +  z  =  6. 


1      1 

-+  - 


1  _  1 
X  y 
3      2 


=  4. 


by  +  z  —  4:U 
3y  +  u  =  \2, 
x  +  2y  +  Su  =  8, 


17, 


10. 


2/  +  2  +  u  =  4, 

X  +  Z  +  M  =  3, 

X  +  2/  +  M  =  1, 

X  +  y  +  2  =  10. 

CX  +  &?/  =  I, 

by  +,az  —  m, 

11. 


Ix  =  my  =  nz,  r2x  =  3y  =  6z, 

_ax  +  by  +  cz  =  d.  '   1  (x+22/+2-16){3x-2y  +  20)=0. 

Show  that  the  following  systems  are  not  independent  and  find  the 
identity  which  connects  the  equations  of  each  system. 

3,  fSx-Sy  +  7z  =  lO, 

13. 


2/  -  z  =  -  5, 

14.    }  2x  +  52/-32  =  12, 

2-X  =  2. 

I6x  +  9y-z  =  80 

PROBLEMS 

396  The  following  problems  can  be  solved  by  means  of  simple 
equations  in  two  or  more  unknown  letters,  as  a;,  y,  •••.  How 
many  of  these  letters  it  is  best  to  employ  in  any  case  will 
depend  on  the  conditions  of  the  problem.  But  when  a  choice 
has  been  made  of  the  unknow^n  numbers  of  the  problem 
which  X,  1/,  •  ■  •  are  to  represent,  and  the  remaining  unknown 
numbers,  if  any,  have  been  expressed  in  terms  of  these  let- 
ters, it  will  V)e  found  that  the  conditions  of  the  problem  still 
unused  will  yield  just  as  many  independent  and  consistent 
equations  connecting  x,  ?/,•••  as  there  are  letters  x,  i/,  ■■■.  In 
fact,  if  they  gave  more  than  this  number  of  equations,  the 


SIMULTANEOUS    SIMPLE   EQUATIONS  149 

problem  would  have  no  solution ;  if  less,  an  infinite  number 
of  solutions,  §  395. 

The  remark  of  §  352  on  the  restrictions  which  the  nature 
of  the  problem  may  impose  on  the  character  of  the  unknown 
numbers  applies  here  also. 

Example  1.  In  a  certain  number  of  three  digits,  the  second  digit  is 
equal  to  the  sum  of  the  first  and  third,  the  sum  of  the  second  and  third 
digits  is  8,  and  if  the  first  and  third  digits  be  interchanged,  the  number 
is  increased  by  99.    Find  the  number. 

Let       X  =  hundreds  digit,  y  =  tens  digit,  z  =  units  digit. 

Then  the  number  is  100  x  +  lOy  +  z. 

But,  by  the  conditions  of  the  problem,  we  have 

X  +  z  =  y,  (1) 

2/  +  z  =  8,  (2) 

1002  + 10?/  +  x  =  100a;+ 10?/  +  z  +  99.  (3) 

Solving  (1),  (2),  (3)  we  find  x  =  2,  7/  =  5,  2  =  3. 
Hence  the  number  is  253. 

Example  2.  After  walking  a  certain  distance  a  pedestrian  rests  for 
30  minutes.  He  then  continues  his  journey,  but  at  |  of  his  original 
rate,  and  on  reaching  his  destination  finds  that  he  has  accomplished 
the  entire  distance,  20  miles,  in  6  hours.  If  he  had  walked  4  miles 
further  at  the  original  rate  and  then  rested  as  before,  the  journey  would 
have  taken  5&  hours.  What  was  his  original  rate,  and  how  far  from  the 
starting  point  did  he  rest  ? 

Let  X  =  original  rate  in  miles  per  hour,  and  let  y  =  number  of  miles 
from  starting  point  to  resting  place. 

Expressing  in  terms  of  x  and  y  the  number  of  hours  taken  by  (1)  the 
actual  journey,  (2)  the  supposed  journey,  we  have 

X      2       7x/8         '       ^'  X  2       7x/8        ^        ^' 

Solving  (1),  (2)  for  y/x  and  1/x,  we  find  y/x  =  3/2,  ] /x  =  1/4. 
Hence  x  —  4,  y  =  6. 

Example  3.  Two  vessels,  A  and  B,  contain  mixtures  of  alcohol  and 
water.  A  mixture  of  3  parts  from  A  and  2  parts  from  B  will  contain 
40%  of  alcohol ;  and  a  mixture  of  1  part  from  A  and  2  parts  from  B  will 


150  A   COLLEGE    ALGEBRA 

contain  32%  of  alcohol.     What  are  the  percentages  of  alcohol  in  A  and  B 
respectively  ? 

Let  X  and  y  denote  the  percentages  of  alcohol  in  A  and  B  respectively. 

Then  by  the  given  conditions  we  shall  have 

3x      2,^4^  -  +  ^  =  ^.         (2) 

6         5        100  ^  '  3       3        100  ^  ' 

Solving  (1),  (2)  we  find  x  =  52/100,  or  52%,  y  =  22/100,  or  22%. 

EXERCISE   XI 

1.  Find  three  numbers  whose  sum  is  20  and  such  that  (1)  the  first 
plus  twice  the  second  plus  three  times  the  third  equals  44  and  (2)  twice 
the  sum  of  the  first  and  second  minus  four  times  the  third  equals  —  14. 

2.  The  sum  of  three  numbers  is  51.  If  the  first  number  be  divided 
by  the  second,  the  quotient  is  2  and  the  remainder  5  ;  but  if  the  second 
number  be  divided  by  the  third,  the  quotient  is  3  and  the  remainder  2. 
What  are  the  numbers  ? 

3.  Find  a  number  of  two  digits  from  the  following  data  :  (1)  twice 
the  first  digit  plus  three  times  the  second  equals  37  ;  (2)  if  the  order  of  the 
digits  be  reversed,  the  number  is  diminished  by  9. 

4.  A  owes  $5000  and  B  owes  $3000.  A  could  pay  all  his  debts  if 
besides  his  own  money  he  had  |-  of  B's  ;  and  B  could  pay  all  but  $100 
of  his  debts  if  besides  his  own  money  he  had  \  of  A's.  How  much  money 
has  each  ? 

5.  Find  the  fortunes  of  three  men.  A,  B,  and  C,  from  the  following 
data  :  A  and  B  together  have  p  dollars  ;  B  and  C,  q  dollars ;  C  and  A, 
r  dollars.  What  conditions  nuist  p,  q,  and  r  satisfy  in  order  that  the 
solution  found  may  be  an  admissible  one  ? 

6.  A  sum  of  money  at  simple  interest  amounts  to  $2556.05  in  2  years 
and  to  $2767.10  in  4  years.  What  is  the  sum  of  money,  and  what  the 
rate  of  interest  ? 

7.  A  man  invested  a  certain  sum  of  money  partly  in  4%  bonds  at  par, 
partly  in  5%  bonds  at  1 10,  and  his  income  from  the  investment  was  $650. 
If  the  4%  bonds  had  been  at  80  and  the  5%  bonds  at  110,  his  income  from 
the  investment  would  have  been  $100  greater.     How  much  did  he  invest  ? 

8.  Find  the  area  of  a  rectangle  from  the  following  data :  if  6  inches 
be  added  to  its  length  and  6  inches  to  its  breadth,  tlie  one  becomes  |  of  the 
other,  and  the  area  of  the  rectangle  is  increased  by  84  square  inches. 


SIMULTANEOUS    SIMPLE    EQUATIONS  151 

9.  A  gave  B  as  much  money  as  B  had ;  then  B  gave  A  as  much 
money  as  A  had  left ;  finally  A  gave  B  as  much  money  as  B  then  had 
left.     A  then  had  $16  and  B  $24.     How  much  had  each  originally  ? 

10.  A  and  B  together  can  do  a  certain  piece  of  work  in  5}  days ;  A 
and  C,  in  44  days.  All  three  of  them  work  at  it  for  2  days  when 
A  drops  out  and  B  and  C  finish  it  in  Ij^^  days.  How  long  would  it  take 
each  man  separately  to  do  the  piece  of  work  ? 

11.  Two  points  move  at  constant  rates  along  the  circumference  of  a 
circle  whose  length  is  150  feet.  When  they  move  in  opposite  senses 
they  meet  every  5  seconds  ;  when  they  move  in  the  same  sense  they  are 
together  every  25  seconds.     What  are  their  rates  ? 

12.  It  would  take  two  freight  trains  whose  lengths  are  240  yards  and 
200  yards  respectively  25  seconds  to  pass  one  another  when  moving  in 
opposite  directions  ;  but  were  the  trains  moving  in  the  same  direction,  it 
would  take  the  faster  one  3|  minutes  to  pass  the  slower  one.  What  are 
the  rates  of  the  trains  in  miles  per  hour  ? 

13.  Two  steamers,  A  and  B,  ply  between  the  cities  C  and  D  which  are 
200  miles  apart.  The  steamer  A  can  start  from  C  1  hour  later  than  B, 
overtake  B  in  2  hours,  and  having  reached  D  and  made  a  4  hours'  wait 
there,  on  its  return  trip  meet  B  10  miles  from  D.  What  are  the  rates  of 
A  and  B  ? 

14.  In  a  half-mile  race  A  can  beat  B  by  20  yards  and  C  by  30  yards. 
By  how  many  yards  can  B  beat  C  ? 

15.  A  and  B  run  two  440-yard  races.  In  the  first  race  A  gives  B  a  start 
of  20  yards  and  beats  him  by  2  seconds.  In  the  second  race  A  gives  B  a  start 
of  4  seconds  and  beats  him  by  6  yards.     What  are  the  rates  of  A  and  B  ? 

16.  Two  passengers  together  have  500  pounds  of  baggage.  One  pays 
$1.25,  the  other  $1.75  for  excess  above  the  weight  allowed.  If  the  bag- 
gage had  belonged  to  one  person,  he  would  have  had  to  pay  $4.  How 
much  baggage  is  allowed  free  to  a  single  passenger  ? 

17.  Given  three  alloys  of  the  following  composition  :  A,  5  parts  (by 
weight)  gold,  2  silver,  1  lead  ;  B,  2  parts  gold,  5  silver,  1  lead  ;  C,  3  parts 
gold,  1  silver,  4  lead.  To  obtain  9  ounces  of  an  alloy  containing  equal 
quantities  (by  weight)  of  gold,  silver,  and  lead,  how  many  ounces  of  A,  B, 
and  C  must  be  taken  and  melted  together  ? 

18.  A  and  B  are  alloys  of  silver  and  copper.  An  alloy  which  is  5  parts 
A  and  3  parts  B  is  52%  silver.  One  which  is  5  parts  A  and  11  parts  B  is 
42%  silver.     What  are  the  percentages  of  silver  in  A  and  B  respectively  ? 


152  A   COLLEGE   ALGEBRA 

19.  A  marksman  who  is  firing  at  a  target  500  yards  distant  hears  the 
bullet  strike  2|  seconds  after  he  fires.  An  observer  distant  000  yards 
from  the  target  and  210  yards  from  the  marksman  hears  the  bullet  strike 
2^5  seconds  after  he  hears  the  report  of  the  rifle.  Find  the  velocity 
of  sound  and  the  velocity  of  the  bullet,  assuming  that  both  of  these 
velocities  are  constant. 

20.  A  tank  is  supplied  by  tv70  pipes,  A  and  B,  and  emptied  by  a  third 
pipe,  C.  If  the  tank  be  full  and  all  the  pipes  be  opened,  the  tank  will  be 
emptied  in  3  hours ;  if  A  and  C  alone  be  opened,  in  1  hour;  if  B  and  C 
alone  be  opened,  in  45  minutes.  If  A  supplies  100  more  gallons  a  minute 
than  B  does,  what  is  the  capacity  of  the  tank,  and  how  many  gallons  a 
minute  pass  through  each  of  the  pipes  ? 


PROBLEMS  ILLUSTRATING  THE  METHOD  OF  UNDETERMINED 
COEFFICIENTS 

397  We  proceed  to  consider  one  or  two  simple  problems  relating 
to  the  subject  matter  of  algebra  itself. 

The  inquiry  may  arise  with  regard  to  some  particular  func- 
tion of  the  variables  under  consideration,  Can  this  function 
be  reduced  to  a  certain  specified  form  and,  if  so,  what  are  its 
coefficients  when  reduced  to  this  form? 

The  following  example  will  illustrate  the  method  of  attack- 
ing a  problem  of  this  kind. 

Example.  Can  the  expression  x^  +  4x  +  6  he  reduced  to  the  form  oi 
a  polynomial  of  the  second  degree  in  z  +  1,  and,  if  so,  what  are  its  coef- 
ficients when  reduced  to  this  form  ? 

The  most  general  expression  of  the  form  in  question  may  be  written 
a(x  +  1)2  +  6(x  +  1)  +  c,  where  a,  b,  c  denote  constants. 

Hence,  if  the  reduction  under  consideration  is  possible,  we  must  have 

a;2  +  4x  +  6  =  a(x  + l)2  +  ?,(x  +  l)  +  c  (1) 

or  x"^ +  Ax  +  (j  =  ax^  +  {2  a +  h)x  +  (a  +  b  +  c).  (2) 

By  §  284,  (2)  and  hence  (1)  will  be  an  identity  when,  and  only  when, 

the  coefficients  of  like  powers  of  x  in  (2)  are  equal,  that  is,  when  a  =  1, 

2a  +  b  =  4,  a  +  ?>  +  c  =  6,  or,  .solving  for  a,  b,  c,  when  a  =  l,  b  =  2,c~3 

Hence  x2  +  4x  +  6  =  (x  +  1)2  +  2(x  +  1)  +  3- 


SLMLLTANEOCS    SIMPLE    EQUATIONS  153 

Observe  tliat  we  set  the  given  expression  equal  to  an  expres- 
sion of  the  required  form  but  with  undetermined  coefficients. 
We  then  find,  that  to  make  this  supposed  identity  true,  the 
coefficients  must  satisfy  certain  conditional  equations.  And 
by  solving  these  equations  we  obtain  the  values  of  the 
coefficients. 

The  following  is  a  more  general  kind  of  problem  including 
that  just  considered. 

Certain  conditions  are  stated  and  the  question  is  then 
asked,  Does  any  function  of  a  certain  specified  form  exist 
which  will  satisfy  these  conditions,  and,  if  so,  what  are  its 
coefficients  ? 

To  solve  such  a  problem,  we  construct  an  expression  of  the 
form  in  question  with  undetermined  coefficients.  These  coeffi- 
cients are  the  unknown  numbers  of  the  problem  and  the  given 
conditions  yield  the  system  of  equations  which  they  must 
satisfy.  If  this  system  of  equations  has  a  single  solution,  we 
obtain  a  single  function  satisfying  the  given  conditions ;  if 
the  system  has  no  solution,  no  such  function  exists ;  if  the 
system  has  infinitely  many  solutions,  the  problem  is  indeter- 
minate, there  being  infinitely  many  functions  satisfying  the 
given  conditions.  It  is  here  supposed  that  the  function  under 
discussion  is  definite  expression,  §  264. 

Example.  If  possible,  find  a  polynomial  in  x,  of  the  second  degree, 
which  has  the  value  0  when  x  =  \  and  when  x  =  3,  and  the  value  6  when 
x  =  4. 

The  polynomial  in  question  must  have  the  form  ax^  +  hx  +  c.  And 
by  the  conditions  of  the  problem 

a-|-6  +  c  =  0,        9a  +  3  6  +  c  =  0,         16a  +  46  +  c  =  6. 

Solving  for  a,  6,  c,  we  find  a  =  2,  6  =  —  8,  c  =  6. 
Hence  the  required  polynomial  is  2  x^  _  8  x  +  6. 

Had  the  problem  been  to  find  a  polynomial  of  the^rs^  degree  satisfy- 
ing the  given  conditions,  there  would  have  been  no  solution  ;  had  it  been 
to  find  one  of  the  third  degree,  there  would  have  been  infinitely  many 
somtions. 


154  A    COLLEGE   ALGEBRA 

399  The  method  illustrated  in  the  preceding  sections  is  called  the 
method  of  xmdetermined  coefficients.  It  is  the  principal  method 
of  investigation  in  algebra  and  we  shall  often  have  occasion  to 
apply  it  as  we  proceed. 

EXERCISE  XII 

1.  Express  3  x"  -  x^  +  2  x  -  5  as  a  polynomial  in  x  -  2. 

2.  Express  4x2  +  8x  +  7asa  polynomial  in  2 x  +  3. 

3.  rind/(x)  =  ax2  +  6x  +  c  such  that 

/(-l)  =  n, /(l)  =  -5, /(5)  =  6. 

4.  Find  f(x)  =  ax^  +  ftx^  +  ex  +  d  such  that 

/(O)  =  5,  /(-  1)  =  1,  /(I)  =  9,  /(2)  =  3L 

5.  Find/(x,  y)  =  ax  +  hy  -\-  c  such  that 

/(O,  0)  =  4,  /(4,  4)  =  0,  /(I,  0)  =  6. 

6.  Find  a  simple  equation  ax  +  6y  +  1  =  0  two  of  whose  solutions 
are  X  =  3,  y  =  1  and  x  =  4,  2/  =  -  1. 

7.  Can  a  simple  equation  ax  -\- hy  -\-  c  =  0  be  found  which  has  the 
three  solutions  x  =  3,  y  =  l;  x  =  4,  2/=-l;  x  =  l,  2/  =  l? 

8.  Find  the  simple  equation  whose  graph  is  the  straight  line  deter- 
mined by  the  points  (2,  3),  (-  4,  5). 

9.  Determine  c  so  that  the  graph  of3x  +  2/  +  c  =  0  will  pass  through 
the  point  (-  2,  3). 

10.  Find  two  simple  equations,  ax  +  fey  +  1  =  0  (1),  a'x  +  ft'y  + 1  =0  (2), 
such  that  both  are  satisfied  by  x  =  2,  y  =  3  and  also  (1)  by  x  =  7,  y  =  5 
and  (2)  by  x  =  3,  ?/  =  7. 

Plot  the  graphs  of  these  equations. 

11.  Find  the  equation  x^  +  6x2  +  ex  +  (Z  =  0  whose  roots  are  -  2,  1, 
and  3. 

12.  Find  an  equation  of  the  form  x^  +  hxy  +  ex  +  %  =  0  which  haa 
the  solutions  x=l,  ?/  =  0;  x  =  2,  2/  =  l;  x  =  —  2,  2/  =  l. 

13.  Express  3  x  +  2  ?/  —  3  in  the  form 

a(x  +  2/-  l)  +  6(2x-2/  +  2)  +  c{x  +  2y-3), 
-where  a,  6,  and  c  denote  constants. 


THE    DIVISION    TRANSFORMATION  155 

V.     THE   DIVISION  TRANSFORMATION 

THE   GENERAL   METHOD 

Preliminary  considerations.     In  §  319  we  defined  the  quotient    400 
of  .i  by  B  as  the  simplest  form  to  which  the  fraction  A/ B 
can  be  reduced  by  the  rules  of  reckoning. 

We  are  now  to  give  a  general  method  for  finding  the  quotient 
as  thus  defined,  when  A  and  B  are  polynomials  in  the  same 
letter,  as  x,  and  the  degree  of  A  is  not  less  than  that  of  B. 

1.  It  is  then  possible  that  i?  is  a  factor  of  A,  in  other  words, 
that  A  can  be  reduced  to  the  form 

A  =  QB,  (1) 

where  Q  is  an  integi^al  function  of  x. 

We  then  have  —  =  Q, 

B 

that  is,  the  quotient  of  ^  by  £  is  the  integral  function  Q; 
and  we  say  that  A  is  exactly  divisible  by  B. 

Thus,  ifJ.  =  «3  +  4a;2_2x-5  and  i?  =  x2  +  3  x  -  5,  it  will  be  found 
that  x^  +  4x2  —  2x  —  5  =  (x  +  1)  (x^  +  3x  —  5),  an  identity  of  the  form 
(1),  q  being  x  +  1. 

^      x3  +  4x2-2x-5 
Hence  —  = =  x  +  1. 

B  x2  +  3x-5 

2.  But  it  will  usualli/  happen  that  B  is  not  a  factor  of  A. 
We  cannot  then  reduce  A  to  the  form  QB;  but,  as  we  shall 
show,  §  401,  we  can  reduce  it  to  the  form 

A  =  QB  +  R,  (2) 

where  both  Q  and  R  are  integral  functions  of  x,  and  the  degree 

of  R  is  less  than  that  of  B. 

A  R 

We  then  have  —  =  Q  -\-  —, 


156  A   COLLEGE    ALGEBRA 

that  is,  the  quotient  of  A  by  B  is  the  sum  of  an  integral  func- 
tion, Q,  and  a  fraction,  B/B,  whose  numerator  is  of  lower 
degree  than  its  denominator. 

In  this  case  we  call  Q  the  integral  part  of  the  quotient,  and 
R  the  remainder. 

Thus,  if  J.  =  a;3  +  2  x2  +  3 X  +  3  and  B  =  x2  +  2  X  +  2,  we  can  at  once 
reduce  A  to  the  form  (2)  by  writing  it 

x3  +  2x2  +  3a;+  3  =  x(x2  4-2x  +  2)  +  (x  +  3), 
where  Q  is  x  and  E  is  x  +  3,  which  is  of  lower  degree  than  B. 

„  ^      x3  +  2x2  +  3x  +  3  ,         x  +  3 

^^'^'^^       ^^     X2  +  2X  +  2     =^  +  xm:¥^T^- 

401  The  division   transformation.     It    remains    to   show  how  to 

reduce  A  to  the  form  QB  +  R,  where  R  is  of  lower  degree  than 
B  and  may  be  0.  The  process  by  which  this  is  usually  accom- 
plished is  called  the  division  transformation,  or  "  long  division." 
It  is  illustrated  in  the  following  example. 

Let  .4  =  2  X*  +  3  ic'  +  4  a;2  +  a;  -  2  and  £  =  a-2  -  a;  +  1. 

Here  the  degree  of  B  is  two,  and  the  problem  is  to  find  an 
integral  function,  Q,  such  that  the  remainder,  iJ,  obtained  by 
subtracting  QB  from  A,  shall  be  of  \\\q  first  degree  at  most  and 
may  be  0 ;  for  if  such  a  function,  Q,  be  found,  we  shall  have 
A  -  QB  =  B,  and  therefore  .1  =  QB  +  R. 

Since  the  degree  of  A  is  four  and  that  of  R  is  to  be  not 
greater  than  one,  Q  must  be  such  that  the  first  three  terms  of  A 
are  cancelled  when  we  subtract  QB.  This  suggests  the  follow- 
ing method  for  finding  Q. 


A  =  2x' 

+  3x8  +  4a-2-l-    x-2 
-2x3  +  2x2 

x^- 

x  +  l=B 

2x-'B  =  2x* 

2x^  + 

5x+7=Q 

A-2x'B  = 

5x«  +  2x2+    ^_2 

(1) 

5xB  = 

5:r8-5x2  +  5x 

A-(2x'  +  5x)B  = 

7x2-4.T-2 

(2) 

7B  = 

7a,2_7^^7 

A-i2x^  +  5x  +  7)B  = 

3a:-9 

=  R 

(3) 

THE   DIVISION   TRANSFORMATION  157 

Evidently  we  shall  cancel  the  leading  term  of  A  if  we  sub- 
tract any  multiple  of  B  which  has  the  same  leading  term  that 
A  has.  The  simplest  multiple  of  this  kind  is  2x'^B,  where 
the  multiplier  2  x^  is  found  by  dividing  the  leading  term  of  A, 
namely  2  a'*,  by  the  leading  term  of  B,  namely  x^. 

Subtracting  2x-B  from  A,  as  above,  we  have 

A  -2x^B^5x^  +  2x^  +  x-2.  (1) 

We  may  cancel  the  leading  term  of  the  remainder  (1)  thus 
obtained,  and  with  it  the  second  term  of  .4,  by  a  similar  process. 
The  quotient  of  5  a;^  by  x^  is  5  a; ;  and  multiplying  £  by  5  x 
and  subtracting  we  have 

^  -  (2  a;2  +  5  cc)  5  =  7  a;2  -  4  X  -  2.  (2) 

Finally,  we  shall  cancel  the  leading,  term  of  the  remainder 
(2),  and  with  it  the  third  term  of  A,  by  subtracting  7  B,  where 
the  multiplier,  7,  is  found  by  dividing  7  x^  by  x'^.     The  result  is 

A  -  (2  a--'  +  5  a-  +  7)  B  =  3  X  -  9.  (3) 

The  remainder  (3)  is  of  the  first  degree,  and  we  obtain  it 
by  subtracting  (2  x'^  -\-  5  x  -\-  1)  B  from  .1. 

Hence  the  polynomials  Q  and  R  which  we  are  seeking  are 

Q  =  2x-  -\-^x  +  l  and  /?  =  3 x  -  9. 

And  writing  the  identity  (3)  thus, 

.4  =  (2a-2  +  5  x  +  7)  £  +  (3  a:  -  9), 

we  have  A  in  the  form  QB  +  E,  where  the  degree  of  R  is  less 
than  that  of  B. 

We  therefore  have  the  following  rule  for  finding  Q  and  R 
when  A  and  B  are  given. 

Arrange  hoth  A  and  B  according  to  desceiiding  powers  ofx. 

Divide  the  leading  term  of  A  by  the  leading  term  of  B ;  the 
quotient  will  be  the  first  term  of  Q. 

Multiply  B  by  this  first  term  of  Q,  and  subtract  the  product 
from  A. 


158  A   COLLEGE   ALGEBRA 

Proceed  in  a  similar  inaniier  with  the  remainder  thus  obtained, 
dividing  its  leading  term  by  the  leading  term,  of  B,  and  so  on. 

Continue  the  process  until  a  remainder  is  reached  which  is  of 
lower  degree  than  B.  We  shall  then  have  found  all  the  terms 
ofQ,  and  the  final  remainder  will  be  K  —  QB  or  E,. 

It  is  customary  to  arrange  the  reckoning  in  the  manner 
illustrated  above.  We  can  then  use  detached  coefficients,  as 
in  multiplication. 

Example  1.  Given  vl  =  2  a;5  -  6  x*  +  7  x^  +  8  x^  -  19  x  +  20,  and 
^  =  x2-3x  +  4;  find  Q  and  R. 

2-6  +  7  +  8-19  +  2011  -3  +  4 


2-6+8  2+0-] 


-1  +  8- 
-1  +  3- 

-19 
4 

5- 

5- 

-15+20 
-15  +  20 

Hence  Q  =  2x3-x+5 
and  E  =  0. 


Observe  that  instead  of  tlie  first  remainder,  —1  +  8  —  19  +  20,  we 
write  only  that  part,  —1  +  8—19,  which  is  involved  in  the  next  subtrac- 
tion ;  also  that  the  second  coefBcient  of  Q  is  0,  because  the  tioo  leading 
terms  of  A  are  cancelled  by  the  first  subtraction. 

Example  2.  Given  ^  =  x*  +  2  x^  +  3  x2  +  2  x  +  4,  B  =  x-  +  2x, 
find  q  and  R. 

402  Remarks  on  this  method.  1.  Observe  that  in  the  division 
transformation  each  intermediate  remainder  plays  the  role  of 
a  new  dividend  ;  also  that  if  R^  denote  any  such  remainder,  C, 
the  part  of  Q  already  obtained,  and  (22  the  rest  of  Q,  we  have 

A  =  Q^B  +  7^1  and  R^  =  QnB  +  R. 

2.  The  process  by  which  Q  and  R  are  found  is  not  itself  the 
division  of  A  by  B,  but  a  preliminary  operation  consisting  of 
multiplications  and  subtractions,  the  aim  of  which  is  to  reduce 
A  to  the  form  A  =  QB  +  R.  The  division  of  ,4  by  B  does 
not  occur  until  we  pass  from  the  identity  A  =  QB  -{-  R  to  the 
identity  A/B=Q  +  R/B. 


THE    DIVISION    TRANSFORMATION  159 

At  the  same  time  it  is  customary  to  call  the  operation  by 
which  Q  and  R  are  found  "division,"  and  also  to  call  Q  the 
"  quotient "  instead  of  the  "  integral  part  of  the  quotient "  even 
when  R  is  not  0 ;  and  we  shall  usually  follow  this  practice. 
But  "  dividing  A  by  £ "  does  not  then  mean,  as  in  §  254, 
finding  an  expression  which  multiplied  by  B  will  produce  A, 
but  finding,  first,  what  multiple  of  B  we  must  subtract  from 
A  to  obtain  a  remainder  which  is  of  lower  degree  than  B,  and, 
second,  what  this  remainder  is.     Compare  §  87. 

3.  The  steps  by  which  the  integral  expression  A  is  reduced 
to  the  integral  form  QB  -\-  R  may  be  taken  whatever  the  value 
of  X.  Hence  A  and  QB  +  R  have  equal  values  for  all  values  of 
X,  ereii  those  values  for  which  B  is  eqnal  to  0.  On  the  other 
hand,  neither  A  / B  nor  Q  +  R/B  has  any  meaning  when  B  =  0. 

Thus,  if  A  =x-  +  x  +  1  and  B  =  x  -  1, 

we  find,  by  §  401,     x2  +  x  +  1  =  (x  +  2)  (x  -  1)  +  3,  (1) 

x^  +  X  +  1  3 

and  therefore  — - — i~  =  x  +  2  + •  (2) 

X  —  1  X  —  1  ^  ' 

Here  B  =  0  when  x  =  1.  Substituting  1  for  x  in  (1)  and  (2),  we  have 
3  =  3,  which  is  true,  but  3/0  =  3  +  3/0,  which  is  meaningless. 

4.  The  transformation  of  A  to  the  form  QB  +  R  is  U7iique, 
that  is,  there  exists  but  one  pair  of  integral  functions  Q  and  it 
(of  which  it  is  of  a  lower  degree  than  B)  such  that 

A  =  QB  +  R. 
For  were  there  a  second  such  pair,  say  Q',  R',  we  should  have 
QB  +  R=Q'B  +  R'  and  therefore  {Q  -  Q')  B  =  R'  -  R. 
But  this  is  impossible,  since  R'  —  R  would  be  of  lower  degree  than  B 
but  (Q  —  Q')B  not  of  lower  degree  than  B. 

The  effect  of  multiplying  the  dividend  or  divisor  by  a  constant.     403 

The  following  theorems  will  be  of  service  further  on. 

1.    If  7re  multiply  the  dividend  by  any  constant,  as  c,  we 
multiply  both  quotient  and  remainder  by  c. 
F.or  if  A  =  QB  +  R,  then  cA  =  cQ-B  +  cR. 


160  A    COLLEGE   ALGEBRA 

2.  If  we  multiply  the  divisor  hy  c,  we  divide  the  quotient  by 
c,  but  leave  the  remainder  unchanged. 

For  if  A  =  QB  +  R,  then  A  =  ^-cB  +  E. 

3.  If  ive  multiply  both  dividend  arid  divisor  by  c,  tve  multiply 
the  remainder  by  c,  but  leave  the  quotient  uiichanged. 

For  if  A  =  qB  +  R,  then  cA  =  Q,-cB  +  cR. 

4.  If  at  any  stage  of  a  division  transformation  we  multiply  an 
intermediate  remainder  or  the  divisor  by  c,  the  final  remainder, 
if  changed  at  all,  is  merely  multiplied  by  c. 

This  follows  from  1  and  2  and  §  402,  1. 

The  student  Avill  do  well  to  verify  these  theorems  for  some 
particular  case. 

Thus,  we  may  verify  the  second  theorem  by  dividing  A  =4x2  +  6a;  +  l 
first  by  B  =  2  X  -  1,  and  then  by  2  5  =  4  x  -  2. 

4  +  6  +  112-1  4  +  6  +  114-2 

4-2         12  +  4  4-2         |l  +  2 
8+1  8+1 

8-4        .-.  Q  =  2x  +  4,  8-4        .-.  Q  =  x  +  2, 

5  E  =  5.  5  R  =  b. 

404         Division  by  the  method  of  undetermined  coefficients.     We  may 
also  find  Q  and  R  when  .1  and  B  are  given,  as  follows  : 

Example  1.     Divide  ^  =  2  x<  +  3  x^  +  4  x2  +  x  -  2  by  B  =  x^  -  x  +  1. 

Since  the  degree  of  A  is  four  and  that  of  B  is  Uoo,  we  know  in  advance 
that  the  degree  of  Q  is  two  and  that  the  degree  of  R  is  one  at  most. 

Hence  let         Q  =  CqX^  +  cix  +  C2  and  R  —  dox  +  dj, 

where  such  values  are  to  be  found  for  the  coefiBclents  Cq,  Ci,  c^,  do,  di  that 
we  shall  have 

2X*  +  3X3  +  4X2  +  X  -  2  =  (cox2  +  CiX  +  C2)  (x2  -  X  +  1)  +  cfox  +  di 


CqX*  +  -  Co  j  -r'  +  Co 

+  Ci  I        -  Ci 
+  C2 


x2  +  ci  X  -t-  r2 1 

-Col        +dl\  (1) 

+  do 


-  Co  +  Cl  = 

3, 

Co-  Ci+  C2  = 

4, 

Ci-  C2+do  = 

1, 

C3+di^- 

-2, 

THE    DIVISION    TRANSFORMATION  161 

But  to  make  (1)  an  identity,  we  must  have,  §  284, 
Co=      2. 

•.  Ci  =      3  +  Co  =      3  +  2         =      5. 

,-.  Co  =      4  -  Co  +  Ci  =      4  -  2  +  5  =      7. 

•.  rfo  =      1  -  ci  +  C2  =      1  -  5  +  7  ==      3. 

•.  (ii  =  -  2  -  C2  =  -  2  -  7  =  -  9. 

Hence  Q  =  2x2  +  5x  +  7  and  i?  =  3x  -  9,  as  in  §  401. 
Example  2.    Divide  6x5  + 13x*  -  12x3  -  11x2  + llx-2  by  2x2  +  a;-2. 

Exact  division.  Let  A  and  B  denote  polynomials  in  x  with  405 
literal  coefficients,  and  suppose  the  degree  of  B  to  be  m.  For 
the  division  of  .4  by  B  to  be  exact,  the  remainder  R  must  equal 
0  identically.  This  requires  that  all  the  coefficients  of  R  be  0. 
Since  the  degree  of  R  is  m  —  1,  it  has  7n  coefficients,  §  277,  and 
evidently  these  coefficients  are  functions  of  the  coefficients  of 
A  and  B.     Hence 

In  order  that  a  jjoli/nomial  A  viai/  be  exactbj  divisible  by  a 
polynominl  of  the  vath  deyree,  I>,  tlic  coeffieients  of  A  and  B 
must  satisfy  m  conditions. 

The  following  example  will  illustrate  this  fact. 

Example  1.  For  \Yhat  values  of  a  and  f>  is  x^  +  3x-  +  6x  +  2  exactly 
divisible  by  x-  +  ax  +  1  ? 

We  have  x^  +  3  x2  +  6x  +2  Ix^  +  ax  +  1 

x^  +  ax'  +  X |x  +  (3  -  a) 

(3-a)X'i  + (6  -  l)x      +2 
(3  -  a)x2  +  (3a  -  a'^)x  +  (3  -  a) 
(6  -  1  -  3a  +  a2)x  +  (a  -  1) 

Hence  a  and  h  must  satisfy  the  turn  conditions 

ft_l_3a  +  a2  =  0,     a  —  1  =  0;    whence  a  =  1  and  6  =  3. 
Example  2.     Determine  I  and  m  so  that  2x3  +  3x2  +  te  +  ?n  ^ay  jj^ 
exactly  divisible  by  x-  +  x  —  6. 

Dividend  and  divisor  arranged  in  ascending  powers  of  x.      Let     406 
A  and  B  denote  dividend  and  divisor  arranged  in  ascendiny 
powers  of  x,  and  suppose  that  A  does  not  begin  with  a  term 


162  A   COLLEGE    ALGEBRA 

of  lower  degree  than  B  begins  with.  We  may  then  obtain  an 
integral  expression  for  A  in  terms  by  B  by  the  process  of 
cancelling  leading  terms  described  in  §  401.  If  A  is  exactly 
divisible  by  B,  this  result  is  the  same  as  when  A  and  B  are 
arranged  in  descending  powers ;  but  if  A  is  not  exactly 
divisible  by  B,  the  result  is  entirely  different.  The  following 
examples  will  make  this  clear. 


H-3x  +  3x2  +  x3  1  +  x 

1  -2x+     x2  1  +  X 

1+     X                       1  +2x  + 

x2 

1+     X              l-3x  +  4x- 

! 

2x  +  3x2 
2  X  +  2  x2 

(1) 

-3x+     X2 
-3x-3x2 

(2) 

X2  +  X3 
X2  +  X3 

4x2 

4x2  +  4x3 

0 

-4x3 

From  this  reckoning  it  follows  by  the 

reasoning  of  §  401  that 

l  +  3x  +  3x2  +  x3  = 

:{1+2X 

+       X2)  (1  +  X) 

(1) 

1  -  2  X  +     x2 

:(1   -3X 

+  4x2)(l  +  x)-4x3. 

(2) 

The  result  (1)  is  the  same  as  that  obtained  when  the  given  dividend 
and  divisor  are  arranged  in  descending  powers  of  x.  This  must  always 
be  the  case  when,  as  here,  tlie  division  is  exact,  as  follows  from  §  402,  4. 

But  the  result  (2)  is  entirely  different  from  that  obtained  when  we 
arrange  1  —  2  x  +  x2  and  1  +  x  in  descending  powers  of  x.     We  then  get 

x2  -  2  X  +  1  =  (x  -  3)  (X  +  1)  +  4.  (3) 

Both  (2)  and  (3)  are  true  identities,  but  they  give  us  different  expres- 
sions for  x2  —  2  X  +  1  in  terms  of  x  +  1  and  lead  to  different  expressions 
for  the  quotient  of  x2  —  2  x  +  1  by  x  +  1 ,  namely  : 

l-2x  +  x2  4x3 

=  l-3x  +  4x2 , 

1  +  X  1  +  X 

*i:il^±i=  x-3     4-^-. 

X  +  1  X  +  1 

407  Observe  that  in  the  arrangement  according  to  ascending 
powers  the  degrees  of  the  leading  terms  of  the  successive 
remainders  increase,  and,  except  when  the  division  is  exact, 
the  process  has  no  natural  termination.  By  taking  steps 
enough  we  can  obtain,  as  the  integral  part  of  the  quotient,  a 


THE    DIVISION   TRANSFORMATION  163 

polynomial  of  as  many  terms  and  therefore  of  as  high  a  degree 
as  we  please.     Hence 

If  A.  and  B  denote  jjoli/tioniials  arranged  in  ascending  poivers 
of  X,  A  not  being  exactly  divisible  by  B  nor  beginning  with 
a  term  of  lower  degree  than  B  begins  with,  we  can  reduce  the 
quotient  of  A  by  B  to  the  form 

B-^  +B' 

ivhere  Q'  and  R'  are  integral  functions  arranged  in  ascending 
j\owers  of  x,  Q'  ending  xoith  as  high  a  power  of  x  as  we  please 
and  E'  beginning  with  a  still  higher  poiver. 

If  the  number  of  terms  in  Q!  is  n,  we  call  Q'  the  quotient  of 
A  by  li  to  n  terms,  and  R'  the  corresponding  remainder. 

When  the  value  of  x  is  small  (how  small  will  be  shown 
subsequently)  we  can  make  the  value  of  R' / B  as  small  as 
we  please  by  taking  n  great  enough ;  that  is,  we  can  find  a 
polynomial  Q'  whose  value  will  differ  as  little  as  we  please 
from  that  of  A  / B.  On  this  account  the  polynomials  Q'  are 
sometimes  called  approximate  integral  expressiotis  for  the  frac- 
tion A/B. 

Thus  "dividing"  1  by  1  —  x  to  n  "steps,"  we  obtain 

1  x" 

:. =  1  +x  +  x2  +  ...+x''-i  +  ; 

I  —  X  1  —  X 

If  we  give  x  any  value  numerically  less  than  1,  we  can  choose  n  so 
tliat  1  +  a;  4-  •  •  •  +  X"  — 1  will  differ  in  value  as  little  as  we  please  from 
1/(1  -x).  Thus,  if  x=:  1/3,  then  xV(l  -  x)  =  1/18,  so  that  1  +  x  +  x2 
differs  from  1/(1  — x)  by  only  1/18.  Similarly  1  +  x  +  x-  +  x^  differs 
from  1  /(I  —  x)  by  only  1/54  ;  and  so  on. 

Quotients  to  n  terms  found  by  the  method  of  undetermined 
coefficients.     We  proceed  as  in  the  following  example. 

Example  1.     Find  the  quotient  (3  —  x)/(l  —  x  +  2x2)  to  four  terms. 

3  —  X 

Let  =  ao  +  aix  +  a^x"  +  asx^  +  •  •  • .  (1) 

1  —  X  4-  2  x2 


164  A    COLLEGE    ALGEBRA 

Multiplying  both  members  by  1  —  z  +  2  x^  and  collecting  terms,  we  have 

x3  +  ...  (2) 


ao  +  ai  x  +  02 

x^  +  as 

-  ao      -  ai 

—  a^ 

+  2ao 

+  2ai 

But  to  make  (2)  an  identity,  we  must  have,  §  284, 
ao  =  3, 
ai  — ao  =  — 1,        .-.        Oi  =  — l  +  ao=2. 
02  —  tti  +  2  ao  =  0,  .-.        a2  =  ai  —  2  ao  =  —  4. 

as  —  a2  +  2  ai  =  0,  .-.        as  =  a2  —  2  ai  =  —  8. 

Hence  (3  -  x)  /  (1  -  x  +  2  x2)  =  3  +  2  a;  -  4  a;'^  -  8  x^  +  .  ■  • . 
Example  2.     Find  (2  +  x  +  3  x-)  /  (1  +  x  -  x-)  to  five  terms. 

409  Polynomials  involving  more  than  one  variable.  Let  two  poly^ 
nomials,  A  and  B,  be  given  which  involve  more  than  one  variable. 
Unless  ^4  is  of  lower  degree  than  B  with  respect  to  some  one 
of  the  variables,  it  is  2)ossihle  that  .4  is  exactly  divisible  by 
B,  in  other  words,  that  an  integral  function  Q  exists  such  that 
A  /B  =  Q.  We  may  discover  whether  or  not  this  is  so,  and  if 
it  is  find  Q,  by  first  arranging  both  .1  and  B  as  polynomials 
in  some  one  of  the  variables  and  then  applying  the  method 
of  §  401. 

Example  1.     Divide  x^  +  j/S  ^  2-3  _  3  xyz  hy  x  -\-  y  -\-  z. 
x3  -?,yz.x^  (2/3  +  Z-)  \x  +  (y  +  z) 


x^  +  {y  +  z)x-^ |x2  -(y  +  z)x  +  {yi  -yz  +  z'^) 

-  {y  +  z)x-  -  Syz-x 

-(y  +Z)x2-(y  +  2)2x 

{y'^-yz  +  z-^)x  +  {y3  +  z^) 
{y^-yz  +  z''^)x  +  {y^  +  z^) 

Hence  (x^  +  y^  +  z^  -  3  xyz)  /  (x  +  y  +  z)  =  x"^  +  y"-  +  2-  ~yz-zx-  xy. 
Example  2.     Divide  2  x-  +  5  xy  +  3  y^  _|.  7  ^  +  n  y  -  4  by  x  +  ?/  +  4. 

If  .4  is  not  exactly  divisible  by  B,  this  method  will  yield  an 
expression  for  ./I//?  of  the  form  A/B=  Q  +  R / B,  where  Q 
and  R  are  integral  with  respect  to  the  letter  of  arrangement, 


THE    DIVISION    TRANSFORMATION  165 

and  R  is  of  lower  degree  than  B  with  respect  to  that  letter. 
But  the  form  of  this  exjjression  lu'dl  depend  on  what  choice  is 
made  of  the  letter  of  arrangement. 

Example.     Divide  'ix'^  +  Q  xy  -\-  y'^  hy  2  x  -\-  y. 

(1)  Choosing  x  as  the  letter  of  arrangement,  we  have 

Ax'^  +  'Ixy  |2x  +  2y  Hence 

4xy+2y'^  ^  =2x+2y- 


_    y.  2x  +  7y     •  2x  +  y 

(2)   Choosing  y  as  the  letter  of  arrangement,  we  have 

y2  +  6  yx  +  4  x2  ly  +  2x 

y'^  +  2yx ly  +  4  X  Hence 

42/X  +  4X2  2/2  +  6-(/x  +  4x2  ,   ^  4x2 

4yx  +  8x2  — =  y  +  ix — — • 

_4a;2  y  +  2^  2/  +  2X 

EXERCISE   Xm 

1.  By  the  method  of  §  401  and  using  detached  coefficients,  divide 

6x4-7x3-3x2  +  24x-20by  3x2 +  x- 6. 

2.  Also  3x*  -  2x3  -  32x2  +  66x  -  35  by  x2  +  2x  -  7. 

3.  Also  2  x5  -  5  X*  +  13  x^  -  15  x2  +  22  x  by  x2  -  2  x  +  4. 

4.  Also  4  x^  -  3  x5  +  19  X*  +  2  x3  +  4  x2  -  4  X  +  7  by  x3  -  x  +  5. 

5.  By  the  method  of  undetermined  coefficients,  §  404,  divide 

2x''  -  3x2  +  X  -  5  by  x2  -  3x  +  2. 

6.  Also  2x5  -  3x*  +  x2  -  5  by  x^  -  3x  +  2. 

7.  Given  4  =  Sx^  -  5x2  -  7x  +  12  and  jB  =  3 x2  +  x  -  5,  reduce  A 
to  the  form  A  =  QB  +  R,  where  R  is  of  lower  degree  than  B.  Also 
write  down  the  corresponding  expression  for  A/B. 

8.  Determine  a  and  b  so  that  x*  +  ax'  +  x2  +  te  4-  1  may  be  exactly 
divisible  by  x2  —  2  x  +  1 . 

9.  For  what  values  of  a  and  b  is  (x*  +  2  x''  +  3  x2  +  ax  +  6)  /  (x2  +  3  x  +  5) 
reducible  to  an  integral  expression  ? 

10.    Divide  x^  +  x^  +  x^  +  x  +  1  +  2  (x*  +  x2)  by  x2  +  x  +  1. 


1G6  A   COLLEGE    ALGEBRA 

11.  Divide  2x2  +  5x2/ -3?/2-5x  + 13?/-  12by  x  +  3?/-4. 

12.  Divide  2  a"^  -  b-  -  6 c^-  -  ab  +  ac  +  bbc  hy  2  a  +  b  -  3 c. 

13.  Divide  a^  (6  +  c)  +  b-  {c  +  a)  -  c^-  {a  +  b)  +  abc  by  ab  +  be  +  ca. 

14.  Divide  x*  +  (a  -  3)x3  +  (4  -  a)x2  -  2ax  +  8a  by  x2  -  3x  +  4. 

15.  Divide  8  x^  -  27  j/^  by  2  x  -  3  y,  using  detached  coefficients. 

16.  Also  X*  -  4  x?/3  +  3  2/*  by  X  -  y. 

17.  Also  6a^  +  a^b  -  a^b^  +  11  a%^  -  5a¥  +  4b^  hy  2a"-  -  ab  +  b\ 

18.  Tlie  dividend  being  2  x^  +  xy^  +  2/^  and  the  divisor  being  2  x  +  y, 
find  Q  and  B,  first,  when  x  is  taken  as  "letter  of  arrangement,"  second, 
when  y  is  taken  as  that  letter. 

19.  Arranging  dividend  and  divisor  in  ascending  powers  of  x,  find 
quotient  to  three  terms  and  remainder  when  dividend  is  1  —  3  x  +  5  x"- 
and  divisor  1  +  x  +  3  x^. 

20.  Also  when  dividend  is  1  +  x  +  3  x^  and  divisor  1  —  3  x  +  5  x^. 

21.  By  the  method  of  undetermined  coefficients,  §  408,  find  to  four 
terms  the  quotient  1  /  (1  —  2  x). 

22.  Also  the  quotient  (2  +  3  x  +  4  x^)  /  (1  -  x  +  2  x'^)  to /our  terms. 

SYNTHETIC   DIVISION  AND  THE   REMAINDER  THEOREM 

410  Synthetic  Division.  We  proceed  to  explain  a  very  expedi- 
tious method  of  making  the  division  transformation,  §  401, 
when  the  divisor  has  the  form  x  —  b,  that  is,  is  a  binomial  of 
the  first  degree  whose  leading  coefficient  is  1. 

Consider  the  result  of  dividing  aoX^  +  "i-c^  +  anX  +  Og  by 
X  —  h. 

a^x^  +  aiX^    +  «2a^  +  «3  U  —  ^ 
L  gpa:8  -  a^bx''  \a^x;^  +  (nj>  +  ffi^r  +  jaj'''  +  ^i^^  +  ^2) 

{ajb  +  ai)«^  +  a^x 

(a  J)  +  ax)x^  —  {ay  +  a^b)x 

(a.Jj^  +  ajj  -\-  a^)x  +  a^ 

(a^b^  +  a,h  +  a„)x  -(a,h'  +  aj>^  +  aJ>) 

UqO^  +  Uib'^  +a2b  +  as  =  R 


THE   DIVISION    TRANSFORMATION  167 

The  coefficients  of  Q  and  R  are 

ao,     ttob  +  «!,     aoP  +  a^b  +  a^,     ajj^  +  a-^b'^  +  aj)  +  a^. 

Observe  that  the  first  of  these  coefficients  is  the  leading 
coefficient  of  the_dividend  and  that  the  rest  may  be  obtained 
one  after^the  other  by  the  following  rule : 

Multijdy  the  coefficient  last  obtained  by  b  and  add  the  next 
unused  coefficient  of  the  dividend. 

Thus,  ao&2  +  ai&  +  a2  =  {a^h  +  ai)  6  +  a^, 

and  aolfi  +  axh-  +  a^h  +  as  =  {aoh"  +  aih  +  02)  6  +  as. 

This  rule  applies  whatever  the  degree  of  the  dividend  may 
be.  For  since  the  coefficient  of  the  leading  term  of  the  divisor 
is  1,  each  new  coefficient  of  Q  will  always  be  the  same  as  the 
leading  coefficient  of  the  remainder  last  obtained.  Like  that 
coefficient,  therefore,  it  is  found  by  multiplying  the  preceding 
coefficient  of  Q  by  ft  and  adding  a  new  coefficient  of  the  divi- 
dend. And  for  a  like  reason  we  shall  obtain  R,  if  we  multiply 
the  last  coefficient  of  Q  by  J  and  add  the  last  coefficient  of  the 
dividend. 

Hence,  when  the  divisor  has  the  form  x  —  b  and  the  divi-     411 
dend  the  form  aoX"  +  Oja;""^  -f  •  •  •  +  «„,  we  can  find  Q  and  R 
as  follows,  where  Cq,  c^,  ■■•  c„_i  denote  the  coefficients  of  Q. 


ao      «!        052      •• 

■      «„-i 

a„  \b 

Cpb     cj)     ■  ■ 
Co     Ci      C2 

R 

We  first  write  down  the  coefficients  of  the  dividend  in  their 
proper  order  and  b  at  their  right. 

Under  ao  we  write  Cq,  which  we  know  to  be  the  same  as  ao- 

We  then  multiply  Cq  by  b,  set  the  product  Cob  under  ai,  add, 
and  so  obtain  Cj. 

In  like  manner  we  multiply  c^  by  b,  set  the  product  c^b  under 
a^,  add,  and  so  obtain  c^. 


168  A   COLLEGE   ALGEBRA 

And  we  continue  thus,  multiplying  and  adding  alternately, 
until  all  the  coefficients  Uq,  a^,  ■  ■  ■  a„  are  exhausted. 

Example.     Divide  3  x*  -  5  x^  -  4  x^  +  3  x  -  2  by  x  -  2. 

We  have  3_5-4+3      -£[2 

6         2-4-2 

3         1-2-1,-4 

Hence  Q  =  3  x^  +  x^  -  2  x  -  1  and  E  =  -  4. 

This  very  compact  method  is  called  syntJietic  division.  The 
student  should  accustom  himself  to  employ  it  whenever  the 
divisor  has  the  form  x  —  h. 
412  Remarks  on  this  method.  1.  In  dividing  synthetically  when 
the  dividend  is  an  incomplete  polynomial,  care  must  be  taken 
to  indicate  the  missing  powers  of  a;  by  0  coefficients. 

2.  Since  x  +  b  —  x  —  (—  b),  we  may  divide  synthetically  by 
a  binomial  of  the  form  x  -{-  b.  It  is  only  necessary  to  replace 
b  by  —  b  in  the  reckoning  just  explained. 

Example  1.     Divide  x*  —  1  by  x  +  1. 

Here  x  4-  1  =  x  -  (—1),  and  dividing  by  x  —  (—  1),  we  have 

1+0+0+0      -1    I-  1 

_    ni    ±i    ^     id 
1-1+1-1,         0 

Hence  Q  =  x^-x-  +  x-l  and  R  =  0. 

3.  To  divide  by  a  binomial  of  the  form  ax  —  /?,  write  it 
thus  :   a(x  —  (S/a). 

Then  divide  synthetically  by  a;  —  /3/n-,  and  let  Q  and  R 
represent  the  quotient  and  remainder  so  obtained. 

The  quotient  and  remainder  corresponding  to  the  divisor 
ax-  ft  will  he  Q/a  and  It,  §  403,  2. 

Example  2.     Divide  3  x^  -  11  x^  +  18  x  -  3  by  3  x  -  2. 

Here  3x  —  2  =  3(x  —  2/3),  and  dividing  by  x  —  2/3,  we  have 

3-11+18      -3    [2/3 

J     -_6      _8 

3-9        12,        6 


THE    DIVISION    TRANSFORMATION  169 

Hence  the  required  quotient  is  (3a;2  _  9x -f  12)/3,  or  x*  — 3x  +  4 
and  the  remainder  is  5. 

Example  3.     Divide  5  x^  _  x^  +  x  +  2  by  x  -  3. 

Example  4.     Divide  x^  +  6  X'^  +  11  x  +  6  by  x  +  3. 

Example  5.     Divide  2  x^  -  3  x2  +  8  x  -  12  by  2  x  -  3. 

The  Remainder  Theorem,      When  a  polynomial  in  x  is  divided     413 
by  X  —  h,  a  remainder  is  obtained  which  is  equal  to  the  result  of 
substituting  b  for  x  in  the  dividend  ;  so  that  if  f  (x)  denote  the 
dividend,  f  (b)  will  denote  the  remainder. 

The  demonstration  of  this  theorem  is  given  in  §  410 ;  for  it 
is  there  shown  that  if  we  divide  ao^;^  +  «ia;^  +  a^x  +  a^  by 
X  —  b,  we  obtain  the  remainder  aj)^  +  «ii^  +  cL^b  +  dz,  and,  in 
general,  that  if  we  divide  f(x)  =  ao^^"  +  cfi^""^  +  •  •  •  +  «„  by 
X  —  b,  the  remainder  will  be  Oq^"  +  aji""^  +  •  •  •  +  «„>  ot  f(b). 

The  theorem  may  also  be  proved  as  follows  : 

If /(x)  be  the  dividend,  x  —  &  the  divisor,  0  (x)  the  quotient,  and  R  the 
remainder,  then,  §401, 

/(x)  =  </>  (X)  {x-b)  +  R, 

where  R,  being  of  lower  degree  than  x  —  6,  does  not  involve  x  at  all  and 
therefore  has  the  same  value  for  all  values  of  x. 

The  two  members  of  this  identity  havs  equal  values  whatever  the 
value  of  X.     In  particular  they  have  equal  values  when  z  =  b.     Hence 

f{b)  =  <p{b)(b-b)  +  R. 

But  6  —  6  =  0;  and  since  0  (x)  is  integral,  0  (6)  is  finite. 
Hence  0  (6)  (6  —  6)  =  0,  and  therefore 

/(b)  =  R. 

The  following  example  will  serve  the  double  purpose  of  414 
illustrating  the  truth  of  the  remainder  theorem  and  of  show- 
ing that,  when  b  and  the  coefficients  of  f(x)  are  given  numbers, 
usually  the  simplest  method  of  computing  the  value  of  f(b) 
is  to  divide  f(x)  hy  x  —  b  synthetically  —  the  remainder  thus 
obtained  being  f(b). 


170  A   COLLEGE    ALGEBRA 

Example  1.     What  is  the  value  of 

f{x)  =  5  X*  -  12  x3  -  20  x2  -  43  X  +  6,  when  x  =  4  ? 

1.  By  the  method  of  direct  substitution  we  have 

/(4)  =  5 -4* -12.  43  _  20 -42 -43 -4  + 6  =  1280-768 -320- 172 +  6  =  2& 

2.  By  the  method  of  synthetic  division  we  have 


5 

-  12 

-20 

-43 

+    6 

20 

32 

48 

20 

5 

8 

12 

5, 

26 

Example  2.  Given  /(x)  =  3  x*  -  x3  +  5  x2  -  8  x  +  4.  Find,  by  syn- 
thetic division,  /(2),  /(-  2),  /(4),  /(-  2/3). 

415  Corollary  1 .  7/"  f  (x)  vanishes  when  x  =  b,  then  f  (x)  is 
exactly  divisible  by  x  —  lb,  a?id  conversely. 

For,  by  §  413,  f{b)  is  the  remainder  in  the  division  of  f(x) 
hy  X  —  h,  and  the  division  is  exact  when  the  remainder  is  0. 

Thus,  /(x)  =  x3  -  3  x2  +  2  vanishes  when  x  =  1 ,  for  /(I)  =  1  -  3  +  2  =  0. 
Hence  /(x)  is  exactly  divisible  by  x  —  1,  as  may  be  verified  by  actual 
division. 

Again,  /(x)  =  x"  —  6"  is  exactly  divisible  by  x  —  6,  since  f{b)  =  6"  —  6"  =  0. 

Example  1.  If  x^  +  Sx^  —  m  is  exactly  divisible  by  x  —  2,  what  is  the 
value  of  m  ? 

We  must  have        2^  +  3  •  2^  -  m  =  0,  or  m  =  20. 

Example  2.  Show  that  x"  +  6"  is  exactly  divisible  by  x  +  6  if  n  is  odd, 
but  not  if  n  is  even. 

416  Corollary  2.  If  an  integral  function  of  two  or  more  variables 
vanisJies  when  two  of  these  variables,  as  x  and  y,  are  supposed 
equal,  the  function  is  exactly  divisible  by  the  differeAice  of  these 
variables,  as  x  —  y. 

For  the  function  may  be  reduced  to  the  form  of  a  polyno- 
mial in  X  with  coefficients  involving  the  other  variables.  By 
hypothesis,  this  polynomial  vanishes  when  x  =  y.  It  is  there- 
fore exactly  divisible  hy  x  —  y,  ^  415. 

Thus,  x2  {ij  —  z)  +  2/2  (z  —  x)  +  z2  (x  —  y)  vanishes  when  x  =  y  ;  for  sub- 
stituting y  for  X,  we  have  ij"  {y  -  z)  +  y'^  {z  —  y)  +  z^  {y  ~  y)  ^  0. 


THE   DIVISION    TRANSFORMATION  171 

Hence  x^  {y  —  z)  +  y^  (2  —  ;c)  +  z-  (x  —  y)  is  exactly  divisible  hj  x  —  y. 
We  may  verify  this  conclusion  by  actual  division,  tlius  : 

{y  -  2)x2  -  (y2  _  22)a;  +  {y^z  -  z'^y)\x  -  y 

(y  -  z)x^-  (y2  -  yz)  X  \  {y  -  z)  x  -  {yz  -  z^) 

-{yz-z^)x  +  (2/2z  -z-y) 

-  {yz  -z^)x  +  (y-^z  -  z^y) 

Example.  Show  that  (x  —  y)^  +  {y  —  z)^  +  {z  —  x)^  is  exactly  divisible 
hy  X  —  y,  y  —  z,  and  z  —  x. 

Theorem.     If  a  poli/iiomial  f  (x)  vanishes  when  x  =  a  and  also     417 
ivhen  X  =  b,  then  f  (x)  is  exactly  divisible  hy  (x  —  a)  (x  —  b). 

For  since  /(«)  =  0  by  hypothesis,  f{x)  is  exactly  divisible 
by  a;  —  a,  §  415,  and  if  we  call  the  quotient  ^1  (x),  we  have 

f(x)  =  (x  —  a)  </>!  (x),  where  ^1  (x)  is  integral.  (1) 

If  in  (1)  we  set  x  =  ^,  we  have 

f(b)  =  (b-a)Mb)-  (2) 

But  by  hypothesis  f(b)  =  0,  and  b  —  a  ^  0. 

Therefore,  since  when  a  product  vanishes  one  of  its  factors 
must  vanish,  §  253,  it  follows  from  (2)  that  <^i  (b)  =  0. 

But  if  ^1  (b)  =  0,  then  ^1  (x)  is  exactly  divisible  hj  x  —  b, 
§  415,  and  if  we  call  the  quotient  <i>i(x)  we  have 

<^i  (ic)  =  (a;  —  b)  (f)2  (x),  where  <f>o  (x)  is  integral.  (3) 

Substituting  this  expression  for  (fti^x)  in  (1),  we  have 

f(x)  =  (x-a)(x-b).cl>,(x),  (4) 

which  proves  that  f(x)  is  exactly  divisible  by  (x  —  a)(x  —  b). 

Continuing  thus,  we  may  prove  the  more  general  theorem 

If  f(x)  vanishes  for  x  =  a,  b,  c,  •••,  the7i  f(x)   is  exactly     418 
divisible  by  (x  —  a)  (x  —  b)  (x  —  c)  •  •  •. 

Thus,  2  x3  +  3  x2  -  2  X  -  3  vanishes  when  x  =  l,  for  2  +  3-2-3  =  0, 
and  when  x  =  -  1,  for  -  2  +  3  +  2  -  3  =  0. 

Elence  2x3  +  3x2  —  2x  —  3  jg  exactly  divisible  by  (x  —  1)  (x  +  1),  or 
x2  —  1,  as  may  be  verified  by  actual  division. 


172  A    COLLEGE   ALGEBRA 

Example  L  Find  a  polynomial  /(x),  of  the  second  degree,  which 
will  take  the  value  0  when  x  =  2  and  when  x  =  S,  and  the  value  6  when 
»  =  4. 

Since /(x)  is  of  the  second  degree  and  is  exactly  divisible  by  (x  — 2)(x  — 3), 
§  417,  it  may  be  expressed  in  the  form/(x)  =  ao(x  -  2)  (x  -  3),  where  Uq 
denotes  a  constant. 

And  since /{4)  =  6,  we  have  6  =  ao(4  -  2)  (4  -  3),  whence  Uq  -  3. 

Hence  /(x)  =  3(x  -  2)  (x  -  3)  =  3x2  _  I5x  +  18. 

Example  2.  Find  a  polynomial  /(x)  of  the  third  degree  which  will 
vanish  when  x  =  2  and  when  x  =  3,  and  will  take  the  value  6  when  x  =  1 
and  the  value  18  when  x  =  4. 

Reasoning  as  before  we  have  /(x)  =  (aox  +  ai)  (x  -  2)  (x  -  3)  where 
Oo,  ai  are  constants. 

Again,  since  /(I)  =  6,  and  /(4)  =  18,  we  have 

6  =  (ao  +  ai)  (i  -  2)  (1  -  3),        or       ao  +  ai  =  3,  (1) 

18  =  (4ao  +  ai)(4-2)(4-3),     or    4  ao  +  ai  =  9.  (2) 

Solving  (1)  and  (2),  we  obtain  ao  =  2,  ai  =  1. 

Hence  /(x)  =  (2x  +  1)  (x  -  2)  (x  -  3)  =::  2x3  -  Ox^  +  7x  +  6. 

419  ■'  Theorem.  A  pol>/nomial  £(x);  whose  degree  is  n,  cannot  vanish 
for  more  than  n  values  of  x. 

For  if  f(x)  could  vanish  for  more  than  n  valites  of  x,  it 
■would  be  exactly  divisible  by  the  product  of  more  than  n 
factors  of  the  form  x  —  a,  §  41 8,  which  is  evidently  impossible 
since  the  degree  of  such  a  product  exceeds  7i. 

420  Theorem.  If  we  know  of  a  certain  polynomial  /(x),  ivhose 
degree  cannot  exceed  n,  that  it  will  vanish  for  more  than  n 
values  of  x,  we  may  conchide  that  all  its  coefficients  are  0. 

For  if  the  coefficients  were  not  all  0,  the  polynomial  could 
not  vanish  for  more  than  n  values  of  x,  §  419. 

We  say  of  such  a  polynomial  that  it  vanishes  identically. 

421  Theorem.  If  two  polynomials  of  the  nth  degree,  f  (x)  and 
<^(x),  have  eAjual  values  for  more  than  n  values  of  x,  their 
corresponding  coefficients  are  equal. 


THE   DIVISION   TRANSFORMATION  173 

For  let      f(x)  =  aoa;"  +  ai^;""^  H h  «„ 

and  <^ (cr)  =  h^xf  +  b-ipc"-'^  -\ \-b„\ 

also  let         i//  (x)  =  /(x)  —  ^  (a;) 

=  (ao  -  ^-o) ic"  +  («!  -  ^-i)  aj"-^  +  •  •  •  +  (a„  -  *„). 

Then  yj/  (x)  is  0  whenever  the  values  of  f(x)  and  ^  (x)  are 
the  same,  and  by  hypothesis  these  values  are  the  same  for 
more  than  n  values  of  x. 

Hence  the  polynomial  \p (x)  =  (oq  —  ^o)  ^"  +  •••+(«■„  —  ^„), 
whose  degree  does  not  exceed  n,  vanishes  for  more  than  n 
values  of  x,  and  therefore,  §  420,  all  its  coefficients  are  0. 

Therefore  aQ  — bo  =  0,  a^  — b^  — 0,    ■■•,  a„  —  i„  =0, 

whence  a^  =  b^,  a^  =  J^,  •  •  •,  a„  =  b„, 

that  is,  the  corresponding  coefficients  oif(x)  and  <^  (x)  are  equal. 

Thus,  if  /(x)  =  2 x2  +  6x  +  5  and  0 (x)  =  ax^  +  3x  +  c  have  equal 
values  wheu  x  =  2,  4,  6,  we  must  have  a  =  2,  6  =  3,  and  c  =  5. 

EXERCISE   XIV 

1.  Divide  x*  —  3  x^  —  x^  —  11  x  —  4  by  x  —  4  synthetically. 

2.  Also  5x5  -  6x*  -  8x3  +  7 x2  +  6 X  +  3  by  X  -  3. 

3.  Also  3x*  +  x2  -  1  by  X  +  2. 

4.  Also  3x3+ 16x2  -  13x  -  6  by  3x  +  1. 

5.  Also  3  x3  -  6  x2  +  X  +  2  by  3  X  -  1. 

6.  Also  x3  —  (a  +  6  +  c)  X-  +  {ah  +  ac  -|-  hc)x  —  abc  by  x  —  a. 

7.  Also  2  x^  -  x3y  -  7  x2?/2  +  7  xy3  _  10  y4  by  X  -  2  ?/. 

8.  Given  /(x)  =  2x3  -  5x  +  3.     By  the  method  of  §  414,  find 

/(I),  /(2),  /(5),  /(-  1),  /(-  3),  /(-  6). 

9.  By  aid  of  the  remainder  theorem,  determine  »n  so  that 

x3  +  ?nx2  -  20X  +  6 
may  be  exactly  divisible  by  x  —  3. 

10.  In  a  similar  manner,  determine  I  and  m  so  that  2  x3  —  x^  +  Zx  +  m 
may  be  exactly  divisible  by  (x  +  2)  (x  —  4). 


174  A    COLLEGE    ALGEBRA 

11.  By  §  416,  show  that  Sbm  +  am  —  2an  —  6bn  is  exactly  divisible 
hy  m  —  2  n,  also  by  a  +  3  b. 

12.  By  §§416,  417,  show  that  a{b  -  c)^  +  b{c  -  a)^  +  c{a  -  b)^  is 
exactly  divisible  by  (a  —  b){b  —  c)  (c  —  a). 

13.  Find  the  integral  function  of  x  of  the  third  degree  which  vanishes 
when  a;  =  1,  4,  —  2,  and  takes  the  value  —  16  when  x  =  2. 

14.  Find  the  integral  function  of  x  of  the  third  degree  which  vanishes 
when  X  =  2,  3,  and  takes  the  value  6  when  x  =  0  and  the  value  12  when 
x  =  l. 

15.  Show  that  2  x"*  —  ax -i^- 1  and  x3  +  5x  +  2  cannot  have  equal 
values  for  four  values  of  x. 


EXPRESSION   OF  ONE   POLYNOMIAL   IN   TERMS   OF  ANOTHER 

422         Let  A  and  B  denote  two  polynomials  in  x,  A  of  higher  degree 
than  B. 

Divide  Ahj  B  and  call  the  quotient  Q,  the  remainder  R ;  then 

A  =  QB  +  R.  (1) 

If  Q  is  not  of  lower  degree  than  B,  divide  Qhy  B  and  call 
the  quotient  Qi,  the  remainder  R^ ;  then 

Q=Q,B  +  R,.  (2) 

Similarly,  if  Q^  is  not  of  lower  degree  than  B,  divide  Qi  by 
B  and  call  the  quotient  Q^,  the  remainder  R^;  then 

Qi  =  Q,B  +  /?2-  (3) 

Suppose  that  Q2  is  of  lower  degree  than  B.     We  then  have 

A  =  QB  +  R  by  (1) 

=  \Q,B  +  R,\B  +  R  by  (2) 

=  \(Q,B  +  R,)B  +  R,\B  +  R  by  (3) 

=  QoB^  +  R.B''  +  R^B  +  R, 

where  all  the  coefficients   Q2,  Ri,  Ri,  R  are  of  lower  degree 
than  B. 


THE   DIVISION   TRANSFORMATION  175 

And,  in  general,  if  any  polynomial  A  be  given  which  is 
of  higher  degree  than  B,  and  we  continue  the  process  just 
described  until  a  quotient  is  reached  which  is  of  lower  degree 
than  B,  we  shall  have 

A  =  Q^_iB'-  +  R^_iB'—^  -\ h  R^B  +  R 

where  R,  Ri,  ••-,  Rr-i,  Q,.-i  denote  the  successive  remainders 
and  the  final  quotient,  all  being  of  lower  deyree  than  B. 

Example.     Reduce  x^  _  4  x*  +  3  x^  -  -/  "-^  x  +  4  to  the  form  of  a  poly- 
nomial in  x^  +  X  +  1  with  coefficients  whose  "degrees  are  less  than  two. 
Using  detached  coefficients,  we  may  arrange  the  reckoning  thus  : 


1-4  +  3 
1  +  1  +  1 


-5+2-1 
-5-5-5 

7  +  4  +  1 
7  +  7  +  7 

-3-6+4 
-  3  -  3  -  3 

?1  =  «  - ' 


12  +  3     .-.  Ei  =  12x  + 


-3  +  7     .-.  E  =  -3x  +  7. 
Hence  x^  -  4  x*  +  3  x^  -  x^  +  x  +  4 

=  (x  -  6)  (x2  +  X  +  1)2  +  (12x  +  3)  (x2  +  X  +  1)  -  (3x  -  7). 

In  particular,  this  method  enables  us  to  transform  any  poly-     423 
nomial  in  x  into  a  polynomial  of  the  same  degree  va.  x  —  b  with 
constant  coefficients. 

Example.     Transform  2x^  —  x2  +  4x  —  5  into  a  polynomial  in  x  —  2. 

We  may  perform  the  successive  divisions  synthetically  and  arrange 
the  reckoning  as  follows  : 

2     -    1  +4     -    5[2 

4     6         20 

2+3+10,        15     .-.  E  =  15 

4     14 

2+7,  24                .-.  Bi  =  24 

4 

2,        11  .-.  i?2  =  11  and   Q2  =  2. 

Hence  2  x^  -  x2  +  4  x  -  5  =  2  (x  -  2)^  +  11  (x  -  2)2  +  24  (x  -  2)  +  15. 


176  A   COLLEGE    ALGEBRA 

EXERCISE  XV 

1.  By  the  method  of  §  422  express  x*  +  x^  —  1  iu  terms  of  x^  +  1. 

2.  Also  4  X*  +  2  x3  +  4  x2  +  X  +  6  in  terms  of  2  x'^  +  1. 

3.  Also  2x'^  -  3x6  +  2x5  4-  5x*  -  x2  +  6  in  terms  of  x^  -  x2  +  x  +  3. 

4.  Also  x5  +  x^y^  +  x^y^  +  y^  in  terms  of  x^  +  xy  H-  y"^. 

5.  By  the  method  of  §  423  express  2  x^  -  8  x2  +  x  +  0  in  terms  of  x  -  3 

6.  Also  x5  +  3  X*  -  6  x*  +  2  x2  -  3  X  +  7  in  terms  of  x  +  2. 

7.  Also  x3  +  9  x2  +  27  X  in  terms  of  x  +  3. 

8.  Also  x'  +  3  x2  4-  X  —  1  in  terms  of  x  +  1. 


VI.     FACTORS   OF    RATIONAL   INTEGRAL 
EXPRESSIONS 

PRELIMINARY  CONSIDERATIONS 

424  Factor.  Let  A  denote  a  rational  integral  function  of  one 
or  more  variables.  Any  rational  integral  function  of  these 
variables  which  exactly  divides  A  is  called  2^  factor  of  .-i. 

Hence  in  order  that  a  given  function,  F,  may  be  a  factor  of 
^,  it  is  sufficient  and  necessary 

1.  That  F  be  rational  and  integral  with  respect  to  the 
variables  of  which  ^  is  a  function. 

2.  That  A  be  reducible  to  the  form  A  =  GF,  where  G  also 
is  integral. 

Example  1.  Since  2x2  -  2x?/ =  2x(x  -  ?/),  both  x  and  x-y  are 
factors  of  2  x2  —  2  xy. 

Example  2.  Since  3x2  -  2^2  =  (Vsx  +  v'2y)  ( V3x  -  V22/),  both 
V3  X  +  V2  2/  and  V3  X  —  V2  2/  are  factors  of  3  x2  —  2  ?/2. 

Example  3.  Althouj;h  x  -  y  =  ( Vx  +  Vy)  ( Vx  -  V^),  we  do  not  call 
Vx  +  v^  and  Vx  —  Vy  factors  of  x  —  y,  because  they  are  not  rational 
with  respect  to  x  and  y. 


FACTORS    OF   INTEGRAL    EXPRESSIONS  177 

Note  1.     The  coefficients  of  a  factor  need  not  be  either  integral  or     425 
rational.     On  the  contrary,  they  may  be  numbers  or  expressions  of  any 
kind.     In  Ex.  2  they  are  irrational. 

Therefore,  since  x^  —  y  =  {x  +  \y)  (x  —  V^),  the  expression  x^  —  y, 
regarded  as  a  function  of  both  x  and  y,  cannot  be  factored  ;  but  regarded 
as  a  function  of  x  alone,  it  has  the  factors  x  +  V^  and  x  -  Vy.  And 
the  like  may  be  said  of  other  expressions  which  involve  more  than 
one  letter. 

Note  2.  Except  when  dealing  exclusively  with  functions  having  426 
integral  coefficients,  it  is  not  customary  to  include  a  mere  "numerical 
factor,"  like  2  in  Ex.  1,  in  a  ''st  of  the  factors  of  a  given  integral  func- 
tion, A  ;  for  if,  as  here,  we  do  not  require  the  coefficients  of  an  integral 
function  to  be  integers,  any  mere  number  (or  constant)  whatsoever  may 
be  said  to  divide  A  exactly. 

For  a  like  reason,  if  F  is  a  factor  of  A,  and  c  is  any  constant  (not  0), 
cF  is  also  a  factor  of  A  ;  but  we  regard  F  and  cF  as  essentially  the  same 
factor  and  include  but  one  of  them  in  a  list  of  the  factors  of  A. 

Thus,  in  Ex.  1,  it  would  be  equally  correct  to  say  that  2x  and  x  —  y, 
or  that  —  2  X  and  y  —  x,  are  the  factors. 

Theorem.    If  F  is  a  factor  of  B,  and  ^  is  a  factor  of  A,  then     427 
F  is  a  factor  of  A. 

For,  by  §  424,  A  and  B  are  reducible  to  the  forms 
A  =  GB     and     B  =  HF, 
where  G  and  H  are  integral. 

Hence  A  =  GHF=GH-F, 

that  is,  F  is  a  factor  of  .4,  §  424. 

Prime,  composite.     An  integral  function  may  have  no  other     428 
factor  than  itself  (or  a  constant).     In  that  case  we  call  it  prime. 
But  if  it  have  other  factors,  we  call  it  composite. 

Thus,  X  +  2/2  and  x  —  2  y  are  prime,  but  x^  —  y"-  is  composite. 

A  composite  function,  A,  of  the  nth.  degree,  is  the  product     429 
of  not  less  than  two  nor  more  than  n  prime  functions,  B,  C,  ■  ■  ■. 
These  prime  functions  are  called  the  prime  factors  of  A. 


178  A   COLLEGE    ALGEBRA 

430         In  what  follows  we  shall  assume  that 

1.  Any  given  function  A  has  hut  one  set  of  prime  factors. 

2.  All  other  factors  of  A  are  products  of  these  prime  factors. 

3.  Two  or  more  of  these  pirime  factors  may  be  equal,  but  A 
can  he  expressed  in  only  one  way  as  a  product  of  powers  of  its 
different  prime  factors. 

These  theorems,  of  which  2  and  3  are  corollaries  to  1,  will  be  proved 
in  §§  484,  485  for  the  case  in  which  A  is  a  function  of  a  single  variable, 
and  they  can  be  proved  generally. 

Thus,  since  x^y^  -  2  x^y*  =  xxyyy  (x  -  2  y),  the  prime  factors  of 
x3y3  _  2x2i/*  are  x,  x,  y,y,y,x-2  y.  Its  other  factors,  as  x^,  xy,  and 
so  on,  are  products  of  two  or  more  of  these  prime  factors.  Its  different 
prime  factors  are  x,y,x-2y,  and  it  can  be  expressed  in  but  one  way  as 
a  product  of  powers  of  these  factors,  namely  thus:  x'^y^{x  -  2  y). 

431  Factoring.  To  factor  a  given  function,  A,  completely  is  to 
"resolve  it  into  its  prime  factors,"  that  is,  to  reduce  it  to 
the  form  A  =  BCD--,  where  B,  C,  Z»,  •••  denote  prime 
functions. 

But  ordinarily  we  do  not  attempt  to  discover  these  prime 
factors  at  the  outset.  We  endeavor  first  to  resolve  A  into 
a  product  of  some  ttvo  of  its  factors,  as  F  and  G,  next  to 
resolve  F  and  G,  and  so  on,  until  the  prime  factors  are 
reached.  And  even  the  first  step  in  this  process  may  be 
called  "factoring"  .4. 

Factoring  is  the  reverse  of  multiplication.  A  multiplication  usually 
involves  two  main  steps  :  (1)  a  number  of  applications  of  the  distributive 
law,  in  order  to  replace  {a  +  b)c  by  ac  +  6c,  and  so  on  ;  (2)  the  combina- 
tion of  like  terms  in  the  result  thus  obtained.  To  reverse  the  process, 
we  must  (1)  separate  the  terms  thus  combined  —  it  is  in  doing  this  that 
the  difficulty  of  factoring  consists  —  and  then  (2)  apply  the  distributive 
law  in  order  to  replace  ac  +  be  by  (a  +  b)  c,  and  so  on. 

It  must  not  be  supposed  that  every  composite  function  can  be  actually 
factored.  Thus,  while  it  can  be  proved  that  x&  +  ax^  +  bx-  +  ex  +  d  is 
composite,  it  can  also  be  proved  that  the  factors  of  this  expression  cannot 
be  found  by  algebraic  methods,  that  is,  by  applying,  a  finite  number  of 
times,  the  various  algebraic  operations. 


FACTORS   OF   INTEGRAL   EXPRESSIONS  179 

EXPRESSIONS    WHOSE  TERMS   HAVE  A   COMMON  FACTOR 

Expressions  whose  terms  all  have  a  common  factor,  mono-     432 
mial  or  polynomial,  can  be  factored  by  a  single  application  of 
the  distributive  law,  namely  : 

ab  -{-  ac  -\-  ad  -\-  •  •  ■  =  a  (b  -\-  c  -\-  d  -{-  ■  •  ■). 

Example  1.     Factor  2  a^c  +  2  abc  +  4  ac^  —  6  acd. 

All  the  terms  have  the  factor  2ac.     "  Separating  "  it,  we  have 

2a~c  +  2abc  +  4ac-  -  Gacd  =  2ac{a  -{-  b  +  2c  -  3 d). 
Example  2.     Factor  a{c  —  d)  +  b{d  —  c). 
Both  terms  have  the  factor  c  —  d.     Separating  it,  we  have 

a{c  -  d)  +  b{d  -  c)  =  a(c  -  d)  -  b{c  -  d)  =  {a  -  b)  {c  -  d). 

Factors  such  as  these  should  be  separated  at  the  outset. 

Some  expressions  which  are  not  in  this  form  as  they  stand     433 
can  be  reduced  to  it  by  combining  such  of  their  terms  as  have 
a  common  factor. 

Example  1.     Factor  ac  +  bd  +  ad  +  be. 

Combining  ac  and  ad,  also  be  and  bd,  we  obtain  a(c  +  d)  +  b{e  +  d), 
a  binomial  whose  terms  have  the  common  factor  e  +  d. 

Hence  ae  +  bd  +  ad  +  be  =  {a  +  b)  (c  +  d). 

Observe  that  the  parts  into  which  we  separate  the  expres- 
sion in  applying  this  method  must  all  have  the  same  number 
of  terms. 

jExample  2.     Factor  a"^  -{■  ab  —  bd  —  ad  -\-  ae  —  ed. 

We  must  have  either  two  groups  of  three  terms,  or  three  groups  of 
two  terms.  Four  of  the  terms  involve  a,  namely,  a^,  a6,  —  ad,  ac,  and 
the  remaining  two  involve  d,  namely,  —  bd  and  —  cd.  To  obtain  groups 
which  have  the  same  number  of  terms  we  combine  the  term  —  ad  with 
the  d-terms,  and  have 

a2  +  a6  +  ac  -  od  -  5d  -  cd  =  a  (a  +  6  +  c)  -  d  (a  +  &  +  c) 
=  {a  -  d)  {a -Y  b  +  c). 


180  A   COLLEGE   ALGEBRA 

EXERCISE  XVI 

Factor  the  following  expressions. 

1.    6  z*j/322  _  12  x2y*z  +  8  x2y3.  2.  2  n2  +  (n  -  3)  n. 

3.   ab  -  a  +  b  -  1.  4.  mx  -  nx  -  mn  +  n"^. 

5.    3xy-2x- 122/  +  8.  6.  lOxy  +  52/2  +  6x  +  Sy. 

7.   x^y"^  -  x^ys  +  2  x22/  -  2  X2/2.  8.  x*  +  x^  +  x2  +  x. 

9.    ac  +  6d  -  (6c  +  ad).  10.  a2c  -  dbd  -  abc  +  a^d. 

11.    ad  +  ce+bd  +  ae  +  cd  +  be.  12.  a'^  +  cd  -  ab  -  bd  +  ac  +  ad. 

FACTORS   FOUND  BY  AID   OF  KNOWN   IDENTITIES 

434  In  the  second  chapter  we  derived  a  number  of  special 
products,  as  (a  +  b)  (a  —  b)  =  a-  —  b^.  If  a  given  function,  A, 
can  be  reduced  to  the  form  of  one  of  these  products,  we  can 
write  down  its  factors  at  once.  The  following  sections  will 
illustrate  this  method  of  factoring. 

435  Perfect  trinomial  squares.  This  name  is  given  to  expressions 
which  have  one  of  the  forms  a"^  ±  2  ab  -\-  b'\  Such  expressions 
can  be  factored  by  means  of  the  formulas  : 

a^-\-2ab-\-b^  =  (a  +  b)(a  +  b)  =  (a  +  by. 
a''-2ab  +  b''  =  (a-b)(a-b)  =  (a-  bf. 

Observe  that  in  a  perfect  trinomial  square  (properly 
arranged)  the  middle  term  is  twice  the  product  of  square  roots 
of  the  extreme  terms,  and  that  the  factors,  which  are  equal, 
are  obtained  by  connecting  the  principal  square  roots  of  the 
extreme  terms  by  the  sign  of  the  middle  term. 

To  extract  the  square  root  of  the  perfect  square  is  to  find 
one  of  these  equal  factors. 

Example  1.     Factor  9  x2  -  12  xy  +  4  y^. 

This  is  a  perfect  square,  since  V2xy  =  2  v9x2  •  V4?/2. 

And  since  V9x2  =  3 x,  V4y2  =  2 y,  and  the  sign  of  the  middle  term 
is  -,  we  have  9x2  _  \2xy  +  42/2  =  {3x  -2  2/)(3x  -  2y)  =  (3x  -2yY. 


FACTORS    OF    INTEGRAL    EXPRESSIONS  181 

Example  2.     Factor  a-  +  b"  +  c-  +  2  ab  +  2  ac  +  2bc. 
We  can  reduce  this  to  the  form  of  a  trinomial  square  by  grouping  the 
terms  thus : 

a2  +  2  a6  +  62  +  2  ac  +  2  &c  +  c2  =  (rt  +  6)2  +  2  (a  +  6)  c  +  c2 
=  (a  +  6  +  c)2. 
Example  3.     Factor  the  following  expressions. 
1.    x2  +  14x  +  49.  2.    9 --60  +  02. 

3.    9x'^y^  +  30xy  +  25.  4.    a;2  -  4a;y  +  47/2  4.  6a;  -  12?/ +  9. 

5.   64  a8  -  48  a*  +  9.  6.    a2  +  62  +  c2  -  2  a6  +  2  ac  -  2  6c. 

A  difference  of  two  squares.     Expressions  of  this  form,  or    436 
reducible  to  it,  can  be  factored  by  aid  of  the  formula: 
a^-b-'=(a  +  b)  (a  -  b). 
Thus,     x2  -  2/2  _  22  +  2  2/2  =  x2  -  (2/2  -2yz  +  z"^) 
=  x^-  {y  -  zf 

=  (X  +  y  -  2)  (X  -  2/  +  Z). 

A  very  useful  device  for  reducing  a  trinomial  expression  to 
this  form  is  that  of  making  it  a  perfect  square  by  adding 
a  suitable  quantity  to  one  of  its  terms  and  then  subtracting 
this  quantity  from  the  resulting  expression. 

Thus,  x*  +  x22/2  +  2/*  =  a;*  +  2  ^2^2  +  yi-  x^y^^ 

=   (X2  +  ^2)2   _  y-lyl 

=  (x2  +  2/2  +  a;2/)  {x2  +  y2  _  a;y). 
Example.     Factor  the  following  expressions. 
1.   x*  -  2/6.  2.    6  a3  -  6  a62.  3.    12  aH^  -  1h  axy\ 

4.    25x2" -49x2™.  5.    36x^-1.  6.    x*  -  3 x22/2  +  ?/*• 

A  sum  of  two  squares.     By  making  use  of  the  imaginary     437 
unit  i  =  V—  1,  §§  218,  220,  a  sum  of  squares  a^  +  b"^  can  be 
reduced  to  the  form  of  a  difference  of  squares  and  then  factored 
by  §  436,  the  factors  being  imaginary. 

For  since  %"  =  -!,  we  have  b''  =  -{-  b^)  =  -  (iby. 

Hence  a""  +  b^  =  a^  -  {ibf  =  (a  +  ib)  (a  -  ib). 


182  A   COLLEGE   ALGEBRA 

As  we  have  seen,  §§  219,  220,  i  conforms  to  all  the  ordinai-j 
rules  of  reckoning.     One  has  only  to  remember  when  employ- 
ing it  that  r  =  —  1. 
438         Sums  and  differences  of  any  two  like  powers.     In  §§  308,  309, 
310  we  proved  that 

First.     Whether  n  is  odd  or  eve7i, 

a--b-=^{a-  b)  {a—'  +  a^-^  +  •  •  •  +  ah"~^  +  Z-""')-    (1) 
Second.     When  n  is  even, 

a--b-={a  +  b)  (a"-!  -  a^-'b  +  •■-  +  ab'^~'  -  b"-').    (2) 

Third.     When  n  is  odd, 
«"  +  &"=(«  +  b)  (a"-'  -  a^-H  + ai"-2  +  b^-').    (3) 

Hence  the  following  theorems : 

1.  a"  —  b"  is  always  dunsible  by  2i  —  h. 

2.  a°  —  b"  is  divisible  by  a.  +  h  when  n  is  even. 

3.  a"  +  b"  is  divisible  by  a,  +  10  when  n  is  odd. 

4.  In  every  case  the  quotient  consists  of  the  terms 

^n-i     a"-2b---ab"-2     b"-^ 
connected  by  sigyis  which  are  all  +  when  a  —  b  is  the  divisor, 
but  are  alternately  —  and  +  ivhen  a  +  b  w  the  divisor. 

Thus,   1.  X6  -  1  =  (X  -  1)  (X5  +  X*  +  X3  +  X2  +  X  +  1). 

2.  a;6  -  1  =  (X  +  1)  (x^  -  X*  +  x3  -  x2  +  X  -  1). 

3.  8  a3  +  27  6^c3  =  (2  aY  +  (3  bcf 

=  (2  a  +  3  6c)  [(2  af  -  (2  a)  (3  he)  +  (3  6c)2] 
=  (2  a  +  3  6c)  (4  a2  -  6  a6c  +  9  62c2). 
Example.     Factor  the  following  expressions. 
1.    04x3  -  125  2/\  2.    27  x^  +  1.  3.    16  x*  -  Sly*. 

439         When  n  is  composite.     The  following  theorems  are  an  imme- 
diate consequence  of  §  438,  (1),  (2),  (3)  and  §  436. 

\.    Tfn  is  a  multiple  of  any  integer,  p,  then  a"  —  b"  is  exactly 
divisible  by  aP  —  b^. 


FACTORS    OF   INTEGRAL   EXPRESSIONS  183 

Thus,  x^  -y^  =  (a:2)3  -  (y2)3 

=  (X2  -  y2)  {X*  +  x2?/2  +  yi), 

2.  If  n  is  an  even  multiple  of  any  integer,  p,  then  a"  —  h"  is 
exactly  divisible  by  zP  +  b^. 

Thus,  «6  -  2/6  =  (X3)2  -  (2/3)2 

=  (X3  +  2/3)  (X3  -  2/3). 

3.  7/*  n  is  a?i  odd  multiple  of  any  integer,  p,  then  a"  +  b°  is 
exactly  divisible  by  aP  +  b^,  ivhether  n  itself  is  odd  or  even. 

Thus,  X6  +  2/6  =  (X2)3  +  (2/2)3 

=  (X2  +  2/2)  (X*  -  x22/2  +  2/*). 

4.  7/"  n  zs  a  ^^^ovt^er  o/  2,  ("Ae/i  a"  +  b"  can  be  resolved  into 
factors  of  the  second  degree  by  repeated  use  of  the  device 
exjdained  in  §  43 C. 

Thus,  x8  +  2/8  =  x8  +  2  x42/*  +  2/8-2  x^y* 

=  («*  +  2/*)2-2x-'2/^ 

=  (x*  +  2/*  +  V2  x22/2)  (x4  +  2/*  -  V2  x22/2). 
Again, 

X*  +  y*  +  V2x22/2  =  X*  +  2  x22/2  +  2/*  _  (2  -  V2)x2y2 

=  (x2  +  2/2)-^- (2-V2)x22/2 

=  (x2  +  2/2  +  V2  -  V2x2/)(x2  +  2/2  -  V2  -  V2X2/), 
and  so  on. 

As  each  of  these  "  quadratic  "  factors  can  be  resolved  into  two  (imagi- 
nary) factors  of  the  first  degree  by  §  444,  the  complete  factorization  of 
rt"  +  6"  is  always  possible  when  n  is  a  power  of  2. 

"When  n  is  composite,  it  is  best  to  begin  by  resolving  a"  +  i", 
or  a"  —  b",  into  two  factors  whose  degrees  are  as  nearly  equal  as 
possible.  It  will  always  be  possible  to  factor  at  least  one  of 
the  factors  thus  obtained. 

Thus,  the  factorization  of  x^  —  2/6  given  under  2  is  the  best.  Continu- 
ing, we  have 

X6  -  2/6  =  (X3  +  2/3)  (X3  -  2/3) 

=  (X  +  2/)  (x2  -  X2/  +  2/2)  (X  -  y)  (x2  +  X2/  +  2/2). 


184  A   COLLEGE   ALGEBRA 

Example.     Factor  the  following  expressions. 

1.    X*  +  y^.  2.    x8  -  2/8.  3.    a;9  +  ^9. 

440  The  theorems  of  §§  438,  439  also  apply  to  expressions  of 
the  form  a'"  ±  V'  when  m  and  n  are  multiples  of  the  same 
integer  p. 

Thus,  X6  -  yl5  =  (X2)3  -  (i/5)3 

=  (X2  -  2/5)  (X*  +  x2l/5  +  2/10). 

EXERCISE  XVII 

In  the  following  examples  carry  the  factorization  as  far  as  is  possible 
without  introducing  irrational  or  imaginary  coefficients. 

1.   4  z^y  -  20  x22/2  +  25  X2/3.  2.    28  te'^  -  63  «2/2. 

3.  x2  ^  4  y2  +  9  z2  _  4  a;2/  -  12  2/2  +  6  zx. 

4.  (7  a2  +  2  62)2  _  (2  a2  +  7  62)2. 

B.    (7x2  +  4x-3)2-(x2+4x  +  3)2.         6.    4(1  -  62  -  a6)  -  a2. 

7.    X*  +  x2  +  1.  8.    a*  -  6  a262  +  6<. 

9.    a«  +  4  a2  +  16.  10.    9  x*  +  15  x2y2  +  16  y^. 

11.    4  (a6  +  cd)2  -  (a2  +  62  -  c2  -  cZ2)2. 

12.    570x62/3 -9  2/1^  13.    x»  -  2/9.  14.    xi2  -  2/I2. 

15.    xw  +  2/10.  16.    x5  _  32.  17.    x'  +  2/". 

FACTORS   FOUND   BY  GROUPING  TERMS 

441  Sometimes  the  terms  of  a  polynomial  in  x  can  be  combined 
in  groups,  all  of  which  have  some  common  factor,  as  F.  This 
common  factor  F  is  then  a  factor  of  the  entire  expression. 
Compare  §  433. 

Example  1.     Factor  x^  +  3  x2  -  2  x  -  6. 

Noticing  that  the  last  two  coefficients  are  equimultiples  of  the  first  two, 
we  have 

x8  +  3x2-2x-0  =  x2(x  +  3)  -  2{x  +  3) 

=  (x2  -  2)  (z  +  3)  =  (X  +  V2)  (x  -  V2)  (X  +  3). 


FACTORS   OF   INTEGRAL   EXPRESSIONS  185 

Example  2.     Factor  x^  +  2x^ -\- 2x  +  I. 

Combining  terms  whiich  have  like  coefficients,  we  have 

:=(X2_X+  1)(X  +  1)  +  2X(X  +  1) 

=  (X2  -  X  +  1  i-  2  X)  (X  +  1)  =  {X2  +  X  +  1)  (X  +  1). 

Sometimes  this  can  be  accomplished  by  first  separating  one 
of  the  given  terms  into  two  terms. 
Example  3.     Factor  x^  +  4  x^  +  5  x  +  6. 
We  have 

x3  +  4  x2  +  5  X  +  G  =  x3  +  3  x2  +  x2  +  3  X  +  2  x  +  6 

=  x2  (X  +  3)  +  X  (X  +  3)  +  2  (X  +  3) 
=  (x2  +  X  +  2)  (X  +  3). 
Consider  also  the  following  example. 
Example  4.     Factor  x*  +  2  x^  -f-  3  x^  +  2  x  +  1. 
We  have 
x*  +  2x3  +  3x2  +  2x  +  l  =  x*  +  2x3  +  x2  +  2x2  +  2x  +  l 
=  (x2  +  x)2  +  2  (x2  +  x)  +  1 

=  (X2+X  +  1)2. 

EXERCISE  XVIII 

Factor  the  following  expressions. 

1.   x*-x3  +  x-l.  2.    x^i -x5-8x2  +  8. 

3.    x4- 2x3  +  2x-l.  4.    x'5- 7x2-4x  +  28. 

5.    x6  -  x*y-  -  x-y*  +  ?/.  6.    x^  +  2  x2  +  3  x  +  2. 

yf.    x5  +  2x<  +  3x3  +  3x2  +  2x  +  l.       8.    x*  +  4x3  +  10x2  +  12x  +  9. 

FACTORIZATION   OF  QUADRATIC   EXPRESSIONS 

The  quadratic  x-  +  px  +  q,   factored  by  inspection.     This   is     442 
sometimes  possible,  when  p  and  q  are  integers. 

Since  (x  +  a)(x  +  b)  =  x""  +  (a  +  b)  x  +  ab, 

we  shall  know  the  factors  of  x"^  +  px  +  5-  if  we  can  find  two 
numbers,  a  and  h,  such  that  a  +  I  —  v  and  ab  =  q. 


186  A   COLLEGE   ALGEBRA 

Two  such  numbers  always  exist,  §  444,  though  they  are 
seldom  rational.  But  when  rational  they  are  integers,  §  454, 
and  may  be  found  by  inspection,  as  in  the  following  examples. 

Example  1.     Factor  x2  +  13  x  +  42. 

We  seek  two  integers,  a  and  6,  whose  product  is  42  and  sum  13.  As 
both  ah  and  a  +  6  are  positive,  both  a  and  h  must  be  positive.  Hence  among 
the  positive  integers  whose  product  is  42  —  namely,  42  and  1,  21  and  2, 
14  and  3,  7  and  6  —  we  seek  a  pair  whose  sum  is  13,  and  find  7  and  6. 

Hence  x2  +  isx  +  42  =  (x  +  7)  (x  +  6). 

Example  2.     Factor  x2  -  13  x  +  22. 

Here  both  a  and  h  must  be  negative  ;  for  their  product  is  positive  and 
their  sum  negative.  Hence,  testing  as  before  the  pairs  of  negative  inte- 
gers whose  product  is  22,  we  find  —  11  and  —  2  ;  for  —  11  —  2  =  —  13. 

Hence  x2  -  13x  +  22  =  (x  -  11)  (x  -  2). 

Example  3.     Factor  x2  -  9  x  -  22. 

Here,  since  ah  is  negative,  a  and  h  must  have  opposite  signs ;  and 
since  a  +  6  is  negative,  the  one  which  is  numerically  gi-eater  must  be 
negative.  Hence  we  set  —  22  =  —  22  x  1  =  —  11  x  2,  and,  testing  as 
before,  find  a  =  -  11  and  5  =  2  ;  for  -  11  +  2  =  -  9. 

Hence  x2  -  9  x  -  22  =  (x  -  11)  (x  +  2). 

Example  4.     Factor  the  following  expressions. 
1.    x2 +  3x4-2.  2.    x2-lGx  +  15.  3.   x2-4x-12. 

4.    x2  +  X  -  30.  5.    x2  +  20  X  +  96.  6.    x2  -  21  x  +  80. 

443         The  quadratic  ax^  -f  bx  +  c  factored  by  inspection.     This  is 
sometimes  possible,  when  a,  h,  and  c  are  integers. 

By  multiplying  and  dividing  by  a,  we  may  reduce  ax"^ -{-hx-\-c 
to  the  form  \_(axY  +  h  (ax) -{- ac]  / a,  and  then  factor  the 
bracketed  expression  with  respect  to  ax  by  the  method  just 
explained,  namely,  by  finding  two  integers  whose  product  is 
ac,  and  sum  b. 

Example  1.     Factor  2  x^  +  7  x  +  3. 

Wehave  2xC  +  7x  +  3  =  <^^li±lia^bJ 

^(2x+^)(2x+2)  =  (x  +  3)(2x-fl). 

i 


FACTORS   OF    INTEGRAL    EXPRESSIONS  187 

Example  2.     Factor  abz^  +  {a-  +  b~}x  +  ah. 

ah 
_  {abx  +  a-)  {ahz  +  b") 
~  ^6 

=  (6x  +  a)  (ax  +  h). 

Example  3.     Factor  IQx"^  +  12x-  63. 

In  this  case  it  is  not  necessary  to  multiply  and  divide  by  16,  for  we  have 
16  x2  +  72  X  -  63  =  (4  xf  +  18  (4  x)  -  63 
=  (4x  +  21)(4x-3). 
Example  4.     Factor  the  following  expressions. 

1.    6x2-13x  +  6.  2.    5x2  +  14x-3. 

3.    14  x2  +  X  -  3.  4.    18  x2  +  21 X  +  5. 

5.   49  x2  +  105  X  +  44.  6.    abx"^  -  {ac  -  b^-)  x  -  be. 

The  quadratic  x^  +  px  +  q  or  ax^  +  bx  +  c  factored  by  com-    444 
pleting  the  square.     While  the  preceding  methods  apply  in 
particular  cases  only,  the  following  is  perfectly  general. 

«...     (..?)•=...,..?, 

we  can  make  x^  +  px  a  perfect  square  by  adding  ^^  that  is, 
the  square  of  half  the  coefficient  of  x. 

This  process  is  called  covijiletlng  the  square  of  x^  +  px. 

1,  We  shall  not  affect  the  value  of  x^  -^  px  -\-  q,  if  we  both 
add  and  subtract  jt»^/ 4.  But  by  this  means  we  can  transform 
the  expression  into  the  difference  between  two  squares  and 
then  factor  it  by  §  436.     Thus, 

x"^  +px  +  q  =  x^+px  +  - ■J-'^S' 


=(.+iy-^ 


4        4 


=(..f  +  ^^)(.,|_vS).,, 


188  A   COLLEGE   ALGEBRA 

2.    Since       ax^  +  bx  -{-  c  =  ai  x^  +  - x  -{-  -\f 

we  may  obtain  the  factors  of  this  expression  by  substituting 
b /a  for  2^  and  c/a  for  q  in  (1).  Simplifying  the  result,  we 
have 


'b''-Aac\ 

G  +  lc- 

V62-4«c' 

2a        ) 

2a 

ax-'  +  bx-^c  =  a[  X  +  —  + x  +  -- r——   •  (2) 


Example  1.     Factor  x^  -  6  x  +  2. 

We  have  x2-6x  +  2  =  z2_6x  +  32-32  +  2 

=  (X  -  3)2  _  7 
=  (X  -  3  +  V?)  (X  -  3  -  V7). 

Example  2.     Factor  x2  +  8  x  +  20. 

We  have  x^  +  8  x  +  20  =  x^  +  8  x  +  42  -  42  +  20 

=  (X  +  4)2  +  4 
=  (X  +  4)2  -  4  i2 
=r  (X  +  4  +  2  0  (x  +  4  -  2 1). 

Here  we  first  obtain  a  sum  of  squares,  (x  +  4)2  +  4,  and  then  transform 
this  sum  into  a  difference  by  replacing  4  by  —  4  i-,  §  437.  The  factors 
are  imaginary. 

Examples.     Factor  3x2  —  5x  +  1. 

;x2  -^  5x  +  1  =  3  rx2  _  -X  +  -l 

'5\2     /5\2     r 


We  have 


5\2      13- 
36. 


„/         5       V]3\/         5       Vl3\ 


Example  4.     Factor  the  following  expressions. 

1.   x2  +  10x  +  23.  2.    x2-10x  +  24. 

3.    x2  -  12  X  +  45.  4.    x2  +  X  +  L 

5.   2  x2  +  3  X  +  2.  6.   x2  -  4  ax  -  4  62  -I-  8  a6. 


FACTORS   OF   INTEGRAL   EXPRESSIONS  189 

Homogeneous  quadratic  functions  of  two  variables.    The  methods     445 
of  §§  442-444  are  applicable  to  quadratics  of  the  form 

ax^  +  hxy  +  cy'^. 

Example  1.     Factor  x^  —  8xy  +  Wy'^. 

We  have     x2  -  8 xy  +Uy^  =  x'^-8xy  +  16 1/2  -  2 ?/2 
=  (X  -  4  2/)2  -  2  2/2 
=  [X  -  (4  +  V2)  y]  [X  -  (4  -  V2) y]. 

Example  2.     Factor  the  following  expressions. 

1.    x2  +  5xy  +  4  y~.  2.    x2  —  x?/  +  y^. 

Non-homogeneous  quadratic  functions  of  two  variables.     Such     446 
functions   are  ordinarily  jirime.     But   when  composite,  they 
may  be  factored  as  in  the  following  example. 

Example  1.     Factor  ^  =  x2  +  2x2/  -  8  2/2  +  2  x  +  14  ?/  -  3. 

If  A  is  composite,  it  is  the  product  of  two  polynomials  of  the  first 
degree.  Moreover  its  terms  of  the  second  degree,  x^  +  2xy  —  8  2/2,  must 
be  the  product  of  the  terms  of  the  first  degree  in  these  polynomials. 

We  find  by  inspection  that  x2  +  2  xy  —  8  2/2  =  (x  +  4  ?/)  (x  —  2  y). 

Hence,  if  A  is  composite,  there  must  be  two  numbers,  I  and  m,  such 
that  we  shall  have 

x2  +  2x2/-8  2/2  +  2x  +  14y-3 

=  {x  +  Ay  +  l){x-2y  +  m) 

=  x2  +  2 X2/  -  8 2/2  +  (Z  +  m)x  +  {im-2l)y  +  Im.    (1) 
But  to  make  (1)  an  identity,  we  must  have,  §  285, 

l  +  m  =  2  (2),         -  2  Z  +  4  m  =  14  (3),         Im  =  -  S  (4). 
From  (2)  and  (3)  we  find  I  =  —  I,  m  =  3.     And  these  values  satisfy  (4); 
for  -  1  .  3  =  -  3. 

Therefore  x2  +  2  x?/  -  8  ^2  +  2  x  +  14  2/  -  3  =  (x  +  4  2/  -  l)(x  -  2  ?/  +  3). 

Note.  The  example  shows  how  exceptional  these  composite  functions 
are.  If,  leaving  A  otherwise  unchanged,  we  replace  the  last  term,  —  3, 
by  any  other  number,  A  becomes  prime  ;  for  (4)  will  then  not  be  satisfied 
by  Z  =  -  1,  m  =  3. 

This  method  is  also  applicable  to  homogeneotis  quadratic 
functions  of  three  variables. 


190  A   COLLEGE   ALGEBRA 

Thus,  to  factor  x"^  +  2xy  -  8y'^  +  2xz  +  liyz  -  Sz^,  we  set 
«2  +  2  xy  -  8 1/2  4-  2  xz  +  14  7/z  -  3  22  =  (X  +  4  2/  +  ;«)  (X  -  2  2/  +  mz) 
and  then  proceed  as  above,  again  finding  I  =  —  I,  m  —  S. 
Example  2.     Factor  2  x2  -  7  xy  +  3  ?/  +  5  a;^  _  5  ^/z  +  2  z2. 
Example  3,     Show  that  x2  —  2/2  +  2  x  +  y  —  1  is  prime. 

447  Polynomials  of  the  nth  degree.  We  have  shown  that  every 
polynomial  of  the  second  degree,  a^yx"  +  a^x  +  a^,  is  the  product 
of  factors  of  the  first  degree.  The  like  is  true  of  polynomials 
in  X  of  every  degree,  though  no  general  method  exists  for 
finding  these  factors;  in  other  words, 

Theorem.     Every  ■polynomial  in  x,  of  the  nth  degree, 

f  (x)  =  aox"  +  aiX"-^  -\ h  an_iX  +  a„ 

is  the  product  ofn  factors  of  the  first  degree;  that  is,  there  are 
n  binomials,  x  —  ^1,  x  —  )82,  •  •  •,  x  —  j8^  sucli  that 

f(x)  =  ao(x-/?0(x-A)---(x-^„). 
The  proof  of  this  theorem  will  be  given  later. 

448  Corollary.  A  homogeneous  polynomial  in  two  variables,  x  and 
y,  of  the  nth  degree,  is  the  product  of  n  homogeneous  factors  in 
X  and  y,  of  the  first  degree. 

Thus,  the  homogeneous  polynomial  aox^  +  aix^y  +  02X^2  +  azy^  may 
be  derived  from  aox^  +  aix2  +  a^x  +  as  by  substituting  x/y  for  x  and 
multiplying  the  result  by  y^. 

But  by  §  447,  aoX^  +  aix2  +  otox  +  a.3  =  ao  (x  -  ft)  (x  -  /So)  (x  -  /Sg),  and 
if  we  substitute  x/y  for  x  in  this  identity  and  then  multiply  both  members 
by  y^,  we  obtain 

aox3  +  aix2y  +  a^xy"^  +  03?/^  =  ao(x  -  ^^y)  (x  -  fty)  (x  -  /Ssy). 

EXERCISE  XIX 

Factor  the  following  expressions. 

1.   x2  _  14  jc  +  48.  2.   x2  -  21  X  -  120. 

.3.    6x2_63x-22.  4.    16x2  +  64x  +  63. 

5.    54x2-21x  +  2.  6.    12 x^  +  20 xy  -  8 2/'. 


FACTORS   OF   INTEGRAL   EXPRESSIONS  191 

7.  X*  -  13  x2  +  36.  8.  x-^y  -  3  x2y2  _  18  xy^ 

9.  x2  -  3  X  +  3.  10.  3  x2  +  2  X  -  3. 

11.  x2- 4x2/ -22/2.  12.  x2-6ax-962_i8a6. 

13.  a6x2-(a2  +  62)x-(a2-62).  14.  x2  +  M  +  dx  +  6x  +  cx2  +  cdx. 

15.  x2'-8x;/  +  152/2  +  2x-4?/-3.  16.  x2+3xt/  +  22/2  +  32X  +  5?/2  +  222. 

APPLICATIONS   OF   THE   REMAINDER   THEOREM   AND 
SYNTHETIC   DIVISION 

On  finding  factors  by  aid  of  the  remainder  theorem.     Let  f(x)     44S 
denote  a  polynomial  in  x.     By  the  remainder  theorem,  §  415,  if 
b  denote  a  number  such  that  f(b)  —  0,  then  x  —  b  is  &  factor  of 
f(x).     We  can  sometimes  find  such  a  number  b  by  inspection. 

Example.     Factor /(x)  =  x^  —  5x  +  4. 

Since  /(I)  =  1  —  .5  +  4  =  0,  x  —  1  is  a  factor  of /(x). 
l  +  0-5  +  4[l  Dividing  /(x)   by  x  -  1,   we  obtain   the   quotient 

1 1—  X2  +  X-4. 

1       1  -  4,     0  jjp^^g       ^^^^  =  (X  -  1)  (x2  +  X  -  4). 

Note.     Observe  that  vrhenever,  as  in  this  example,  the  algebraic  sum 
of  the  coefficients  of /(x)  is  0,  x  —  1  is  a  factor  of /(x). 

Polynomials  with  integral  coefficients.     If  asked  to  factor  a    450 
polynomial,  /(a*),  with  integral  coefficients,  it  is  usually  best 
to  look  first  for  any  factors  of  the  first  degree  with  mtegral 
coefficients  that  it  may  have.     These  may  always  be  found  by 
aid  of  the  following  principles,  §§  451,  452. 

A  polynomial  f(x)  =  aoX"  +  ais;""'  +  •  •  •  +  «„,  with  integral     451 
coefficients,  may  have  a  factor  of  the  form  x  —  b,  where  b  is  an 
integer.     But  if  so,  b  must  be  a  factor  of  a„,  the  constant  term 
o/f(x). 

Thus,  let  f{x)  =  OoX^  +  aix2  +  UnX  +  as.     If  x  -  6  is  to  be  a  factor  of 
/(x),  we  must  have,  §  415, 

f{b)  =  aob^  +  ai62  4-  a^b  +  as  =  0, 
and  therefore  (aob^  +  aib  +  02)  &  =  -  as. 

Therefore,  since  ao62  +  aib  +  02  denotes  an  integer,  6  is  a  factor  of  03. 


192  A   COLLEGE    ALGEBRA 

Hence  all  such  factors  x  —  b  may  be  found  as  in  the  follow- 
ing example. 

Example.     Factor  /(x)  =  3  a;^  -  3  x*  -  13  x^  -  1 1  a;2  _  iq  x  -  6. 
The  factors  of  the  constant  term,  —  6,  are  ±1,  ±2,  ±3,  ±6,  and  b 
must  have  one  of  these  values  if  x  -  6  is  to  be  a  factor  of  f(x).     We  test 
these  values  of  b  as  follows  by  synthetic  division. 

3_3_13_11_10  —  61— 1  Since  /(I)  ?i  0,  x  —  1  is  not  a  factor. 

-3+    6+    7+    4  +  6  We  therefore  begin  by  testing  X- (- 1) 

3  3~6^^    T^    4^    6       01—1     <5r  ^  +  1-      The   division  proves  to  be 

_3,     g_    24.    6*         exact,    the    quotient    being    Qi  =  3 x* 

3  ^Tq  II    2  '^    6        0 13  —  6  x^  —  7  x2  —  4  X  —  6,  and  the  remainder 

9        Q  _(.    g'  0.     Hence  x  +  1  is  one  factor  of  /(x), 

3      Q        2        0  ^"^^  ^^  '^  ^^^  product  of  the  remaining 

factors. 

We  have  next  to  factor  Qi.     It  also  proves  to  be  exactly  divisible  by 

X  +  1,  the  quotient  being  Q2  =  Sx'  -  Ox^  +  2x  -  6. 

To  factor  Q2,  vi^hose  constant  term  is  also  —  6,  we  test  successively 
X  +  1,  X  —  2,  X  +  2,  but  in  each  case  obtain  a  remainder  which  is  not  0. 
Hence  none  of  these  are  factors.  But  testing  x  —  3,  we  find  that  it  divides 
Q2  exactly,  the  quotient  being  Q3  =  Sx^  +  2.     Therefore 

/(x)  =  (x+l)2(x-3)(3x2  +  2). 

452         A  polynomial  f{x)  =  a^""  +  ayx''-^  H 1-  o,„  with  integral 

coefficients,  may  have  a  factor  of  the  form  ax  —  /?,  where  a 
and  /?  denote  integers  which  have  no  common  factor.  But 
if  so,  a  must  be  a  factor  of  ao,  and  (3  a  factor  of  a„.  This 
theorem  includes  that  of  §  451. 

Thus,  let  /(x)  =  aox'  +  aiX^  +  a^x  +  03.  If  ax  -  ;3,  or  a{z  -  /3/cr), 
is  to  be  a  factor  of /(x),  we  must  have,  §  415, 


K',) 


8\  efl  /32  B 


and  therefore  aoiS^  +  ai/32a  +  a2/3a2  +  aaa^  =  0.  (1) 

From  (1)  we  obtain  aojS'  =  -  (ai/32  +  oaiSa  +  aza"^)  a.  (2) 

Therefore,  since  a-[^  +  OnPa  +  030-2  is  an  integer,  a-  is  a  factor  of  ao,8^. 

But  a  has  no  factor  in  common  with  /3'',  §  492,  2.     Hence  a  is  a  factor 

of  ao,  §  492,  1. 

Again  from  (1),      {ao^  +  a^^a  +  a,2a2)^  =  -  aacr^,  (3) 

whence,  reasoning  as  before,  we  conclude  that  ^  is  a  factor  of  Og. 


FACTORS    OF    INTEGRAL    EXPRESSIONS  193 

Hence  all  siich  factors  ax  —  /3  niay  be  found  as  in  the 
following  example. 

Example.     Factor  f(x)  =  6  x*  +  5  x'  +  3  x2  -  3  x  -  2. 

If  crx  —  /3  is  to  be  a  factor  of  /(x),  a  must  have  one  of  the  values  ±  1, 
±2,  ±3,  ±6,  and  /3  one  of  the  values  ±1,  ±  2  ;  therefore  /3/a:  must  have 
one  of  the  values  ±1,  ±2,  ±1/2,  ±  1/3,  ±  2/3,  ±  1/6. 

We  may  test  az  —  ^  for  these  various  values  of  ^/a  by  dividing  /(x) 
by  X  —  /3/a  synthetically.  If  the  division  is  exact  and  Q  denotes  the 
quotient,  then  ax  —  ^  is  a  factor  of  f{x)  and  Q/a  is  the  product  of  the 
remaining  factors,  §  412,  3. 

Testing  x  —  1,  x  +  1,  x  —  2,  x  +  2,  successively,  we  find  that  none  of 

them  divides  /(x)  exactly.      But  x+  1/2 

6  +  5  +  3-3-  2-|-  1/2     ^^gg^    ^^^    quotient    being   Qi  =  6  x^  +  2  x^ 

-  -^^^-^  — —  — +  2  X  —  4.     Hence  2  x  +  1  is  one  factor  of 

"*"  "  "^      ~    '  fix)  and  the  product  of  the  remaining  factors 

rf  +  i  +  i       -[-/^  is  Qi/2  =  3x='  +  x2  +  x-2. 

?±^±-?  We   next   proceed   to   factor    Qi/2.      If 

"^      "^    '  ax  — /3  is  to  be  a  factor, /3/(:ir  must  have  one 

■^  "^  ^  +  ^  of  the  values  ±1,  ±2,  ±1/3,  ±  2/3.    But 

we  already  know  that  x  —  1,  x  +  1,  x  —  2,  x  +  2  are  not  factors,  since 
they  are  not  factors  of  f{x).  Testing  x  —  1/3,  x  +  1/3,  we  find  that 
neither  of  them  divides  Qi/2  exactly;  but  x  — 2/3  does,  the  quotient 
being  Q2  =  3x-  +  3x  +  3.  Hence  3x  —  2  is  a  factor  of  Qi/2,  and  the 
product  of  the  remaining  factors  is  (^.2/ 3  =  x-  +  x  +  1.     Therefore 

/(x)  =  (2x+  l)(3x-2)(x2  +  x+  1). 

Note.     It  often  becomes  evident  before  a  division  by  x  —  b  or  x  —  ^/a     453 
is  completed  that  the  division  cannot  be  exact. 

Thus,  the  reckoning  here  given  suffices  to  prove  that  x  —  2  will  not 

divide  5x^  —  ix^  +  x  +  8   exactly ;   for  since   the 

~        +      +     L         "divisor "  2,  the  last  coefficient  of  Q already  found, 

namely  6,  and  the  unused  coefficients  of  the  divi- 

5        6 

dend,  namely  1  and  8,  are  a\\  positive,  the  remaining 

coefficients  of  Q  and  R  must  be  positive,  that  is,  E  cannot  be  0, 

Similarly  we  may  conclude  from  the  reckoning 

^  +  '^  +  '^  -  ^  [1/3.     here  given  that  x  -  1  /3  will  not  divide  Sx^  +  x2 

-  — -  +  X  —  2  exactly.     For  the  number  which  occurs 

next  in  the  reckoning,  namely,  2  •  1  /3,  or  2/3,  is  a 
fraction,  and  this  will  cause  the  remaining  coefficients  of  Q  and  R  to 
be  fractional,  so  that  R  cannot  be  0. 


194  A   COLLEGE   ALGEBRA 

454  It  follows  from  §  452  that  a  ^polynomial  f  (x)  =  x"  H \-  a^, 

whose  leading  coefficient  is  1,  the  rest  being  integers,  cannot 
vanish  for  a  rational  fractional  value  of  x. 

For  if /(/3/a)  -  0,  then/(x)  must  be  exactly  divisible  by  ax  -  ^  pnd 
therefore  a  must  be  a  factor  of  1,  which  is  only  possible  when  cr  =  j^,  1. 

455  Factoring  polynomials  and  solving  equations.  It  follows  from 
§  350  that  the  problem  of  resolving  a  polynomial  f{x)  into  its 
factors  of  the  first  degree  is  essentially  the  same  as  that  of 
solving  the  equation  f{x)  =  0. 

Example  1.     Solve  f{x)  =  2  x*  +  z^  -  17  x'^  -  16  x  +  12  =  0. 
By  §  452  we  find  that  /(x)  =  (2  x  -  1)  (x  +  2)2  (x  -  3). 
Hence  the  equation  /(x)  =  0  is  equivalent  to  the  four  equations 
2x-l  =  0,  x  +  2  =  0,  x  +  2  =  0,  x-3  =  0. 

Therefore  the  roots  of  f(x)  =  0  are  1  /2,  -  2,  -  2,  and  3. 
■  Example  2.     Solve  x^  +  3  x2  =  10  x  +  24. 
Transposing,  x^  +  3  x2  -  10  x  -  24  =  0. 

Factoring,  (x  +  2)  (x  -  3)  (x  +  4)  =  0. 

Hence  the  required  roots  are  —  2,  3,  and  —  4. 

EXERCISE  XX 

Factor  the  following  expressions. 

1.   x3-7x  +  6.  2.   x3  +  6x2  +  llx  +  6. 

3.   x4  -  10x3  +  35x2 -50x  + 24.       4.    x*-2x2  +  3x-2. 

5.    6x^- 13x2-14x-3.  6.    2x3 -5x2y- 2x^2 +  22/8. 

7.    2x*-x3-9x2  +  13x-5.  8.    4x6  -  41  x"  +  46x2  -  9. 

9.    6x5  +  19x*  +  22x3  +  23x2  ^  i6x  +  4. 
10.    5x6  -  7  x5  -  8x*  -  x3  +  7  x2  +  8  X  -  4. 
Solve  the  following  equations. 

11.   x2-4x-12  =  0.  12.   6x2-7x  +  2  =  0. 

13.   x2_5x  =  14.  14.   x2  +  6x  =  2. 

15.   x8-9x2  +  26x  =  24.  16.    x*  +  2x3  -  4x2  -  2x  +  3  =  0 

17.   x«-l=0.  18.    10x3-9x2-3x4-2  =  0. 


FACTORS   OF   INTEGRAL   EXPRESSIONS  195 

EXERCISE  XXI 

The  following  expressions  can  be  factored  by  methods  explained  in  the 
present  chapter.  Carry  the  factorization  as  far  as  is  possible  without 
introducing  irrational  or  imaginary  coefficients. 

1.  6  X2/  +  15  X  -  4  2/  -  10.  2.  a'^bc  -  ac^-d  -  abH  +  bcd^. 

3.  a3  (a  -  6)  +  b^  {b  -  a).  4.  a^  -  81  abK 

5.  a*b  -  aP-b^  +  a^fia  _  a6*.  6.  3  abx'^  -Qaxy  +  bxy  -  2  2/2, 

7.  3x6-192  2/6.  8.  (x2  +  x)3-8. 

9.  64x62/3-2/^5.  10.  x2-(a  -  &)x-a6. 

11.  x2«_3a;«_18.  12.  a;_a;2  +  42. 

13.  3x«  +  3x3-24x-24.  14.  x^  -  9x3  +  8x2  -  72. 

15.  2  xc  -  a2  +  x2  -  2  a6  +  c2  -  62,  iq  ^i  ^^z  _  qq)  +  64. 

17.  a2  -  2  a&  +  62  -  -  5  «  +  5  6  +  G.  18.  x*  -  10  x"y'-  +  9  y*. 

19.  6x2  -  7x2/- 52/2 -4x- 22/.  ,  20.  x*  -  (a2  +  62)  x2  +  a262. 

21.  4(XZ  +  M2/)2-(x2-2/2  +  22_u2)2.  22.  14  x2  +  19  X  -  3. 

23.  1  +  19  y  -  66  2/2.  24.  x2/3  +  55  x:^y^  +  204  x^y. 

25.  a*  -  18  a262c2  +  81  b*c^.  26.  (x2  -  7  x)2  +  6  x2  -  42  x. 

27.  8  (X +  ?/)''- 27  (X  -  2/)3.  28.  (x  -  2  2/)  x^  -  (2/ -  2  x)  j/S. 

29.  x2  +  a2  -  6x  -  a6  +  2  ax.  30.  x^  -  2/^  -  (x  -  2/)^. 

31.  x5  -  x4  -  2  x3  +  2  x2  +  X  -  1.  32.  6-'  +  62  +  1. 

33.  2  x2  +  7  X2/  +  3  2/2  +  9  X  +  2  2/  -  5.  34.  a*  +  4. 

35.  x2-X2/-22/2+4xz-52/z  +  3z2.  36.  4a*  +  3  a262  +  96*. 

37.  x2  -  8  ax  -  40  a6  -  25  62.  38.  x^  +  x*  +  1. 

39.  (x2  +  2x-l)2-(x2-2x  +  l)2.  40.  (ox  +  62/)2  -  (6x  +  ay)2, 

41.  x3  -  ax2  -  62x  +  a62.  42.  x*  +  6x3  _  aH  -  a^b. 

43.  a2-962+  126c-4c2.  44.  8a3  +  12  a2  +  6a  +  1. 

45.  X*  -  2  x3  +  3  x2  -  2  X  +  1.  46.  (ax  +  byf  +  {bx  -  ay)^. 

47.  4x5+4 x*-37x3-37x2+9x+9.  48.  x*-4x  +  3. 

49.  x2  +  5ax  +  6a2-a6-62.  50.  15x3  +  29x2  -  8x  -  12. 

51.  a6cx2  +  (a262  +  c^)  x  +  abc.  52.  2  x^  -  ax2  -  5  a*x  -  2  a^. 


196  A   COLLEGE    ALGEBRA 

53.   (a  -  5)  x2  4-  2  ax  +  («  +  b).  54.    x^^  -  y'^\ 

55.   X*  -6x^  +  1  X-  +  6x  -8.  56.    4 x^  -  3 x  -  1. 

57.  3x5  -  lOx*  -  8x3  -  3x2  +  lOx  +  8. 

58.  5x*  + 24x3-15x2- 118x  + 24. 

59.  a'^bc  +  ac-  +  acd  -  abd  -  cd  -  d"^. 

60.  X*  +  y^  +  z*  -2  x-y-  -  2  yH'^  -  2  zH'^. 

VII.     HIGHEST   COMMON   FACTOR   AND 
LOWEST    COMMON    MULTIPLE 

HIGHEST   COMMON  FACTOR 

456  Highest  common  factor.  Let  .1 ,  B,  •  ••  denote  rational,  inte- 
gral functions  of  one  or  more  variables,  as  x,  or  x  and  y. 

If  A,  B,  ■  ■  ■  have  no  factor  in  common,  we  say  that  they  are 
prime  to  one  another.  But  if  tliey  have  any  common  factor, 
they  have  one  whose  degree  is  highest ;  we  call  it  their  highest 
common  factor'  (H.C.F.). 

Thus,  x2  +  2/2  and  x  +  ?/  are  prime  to  one  another. 
But  4x2/z5,  8x2*,  and  'kx'^yz^  have  the  common  factors  x,  z,  z"^,  z',  xz, 
xz2,  xz^,  and  their  highest  conunon  factor  is  xz^. 

457  Notes.     1.    We  here  ignore  common  numerical  factors. 

2.  It  is  sometimes  said  of  two  or  more  functions  which  are  prime  to 
one  another  that  their  H.C.F.  is  L 

3.  The  numerical  value  of  the  H.C.F.  of  A  and  B  is  not  necessarily 
the  greatest  common  divisor  of  integral  numerical  values  of  A  and  B. 
Thus,  the  H.C.F.  of  (2x  +  l)x  and  (x  -  l)x  is  x.  But  when  x  =  4,  the 
values  of  (2  X  +  1)  x  and  (x  —  1 )  x  are  36  and  12,  and  the  greatest  common 
divisor  of  3()  and  12  is  not  4,  but  12. 

458  Theorem  1 .  The  H.C.F.  o/"  A,  B,  •  •  ■  is  the  product  of  all  the 
different  common  prime  factors  of  K,  B,  •••,  each  raised  to  the 
lowest  2wwer  in  which  it  occurs  in  any  of  these  functions. 

The  truth  of  this  theorem  is  obvious  if  we  suppose  each 
of  the  functions  A,  B,---  expressed  in  the  form  of  a  product  of 


HIGHEST    COMMON    FACTOR  197 

powers  of  its  different  prime  factors  and,  as  in  §  430,  assume 
that  there  is  but  one  such  expression  for  each  function. 

Thus,  the  different  common  prime  factors  of  xyz^,  xz*,  and  x^yz^  are 
X  and  z,  and  the  lowest  powers  of  x  and  z  in  any  of  these  functions  are  x 
and  zK     Hence  the  H.C.F.  is  xz^ 

Observe  tliat  if  it  were  possible  to  express  a  given  function  in  more 
than  one  way  in  terms  of  its  prime  factors,  the  process  described  in  the 
theorem  might  lead  to  various  results  corresponding  to  the  various  ways 
of  expressing  A,  B,  ■  •  -,  and  there  might  be  more  than  one  common  factor 
of  highest  degree. 

Applications  of  this  theorem.     When  the  given  functions  can     459 
be  completely  factored,  their  H.C.F.  may  be  written  down  at 
once  by  aid  of  the  theorem  of  §  458. 

Example  1.     Find  the  H.C.F.  ofx^y"^  -6x*y^  +  9x^y^  and  x*y  -  9xV- 
We  have  x^y"^  -  6  x*?/3  +  9x^y*  =  x^y^  (x  -  3  y)^ 

and  x*y-9x^y^  =  x^y{x-Sy){x  +  3y). 

Hence  the  H.C.F.  is  x^y  (x  -  3 y). 

Example  2.     Find  the  H.C.F.  of  the  following. 

1.  2x'^y-z^,  Sx^y%  and  ix^y*. 

2.  x^  —  y'^,  x^  +  2  xy  +  y-,  and  x^  +  y^. 

3.  x2  -  X  -  6,  x2  +  6  X  +  8,  and  x^  +  5  x  +  6. 

4.  x3-6x2  + llx  -  6  and  2x3  -  9x2  +  7x  +  6. 

If  the  prime  factors  of  one  of  the  functions  A,  B,  •••  are    460 
known,  we  can  find  by  division   or  the  remainder  theorem 
which  of  them,  if  any,  are  factors  of  all  the  other  functions. 
The  H.C.F.  may  then  be  obtained  by  aid  of  §  458. 

Example  1.     Find  the  H.C.F.  of 

/(x)  =  x2  -  3  X  +  2  and  0  (x)  =  x*  -  3  x^  +  5  x2  -  8  x  +  5. 
By  inspection  we  have  f{x)  =  (x  -  1)  (x  —  2).     Testing  x  =  1  and  x  =  2 
in  ^(x),  we  find  0  (1)  =  0,  but  <p{2)  jt  0.     Hence  the  H.C.F.  is  x  -  1. 

Example  2.     Find  the  H.C.F.  of 

/(x)=:x2  +  4x  +  4     and     </.(x)  =  x*  +  5x3  +  9x2  +  8x  +  4. 
Since  /(x)  =  (x  +  2)2,  we  mu.st  find  not  only  whether  x  +  2  is  a  factor 
of  ^  (x),  but  whether  it  is  a  factor  once  or  twice.     Dividinjj  c&  /xi  bi^  ^  -J-  2 


198  A   COLLEGE   ALGEBRA 

(synthetically)  we  obtain  Qi  =  i3  +  3x2  +  3x  +  2  and  Ri  =  0  ;  dividing 
Qi  by  X  +  2  we  obtain  Q2  =  a;^  +  x  +  1  and  E2  =  0.  Hence  the  H.C.F. 
of/(x)  and0(x)is  (x  +  2)2. 

Example  3.     Find  the  H.C.F.  of  the  following. 

1.  a;2  +  X  -  6  and  2x3  +  7x2  +  4x  +  3. 

2.  x2  +  5x  +  6  and  x*  +  6x3  + 13x2  +  16x  +  12. 

3.  (X  -  l)2(x  -  3)3(3x  +  1)2  and  x*  -  5x3  +  x2  +  21x  -  18. 

461  Theorem  2.  Let  A  and  B  denote  two  given  integral  functions, 
and  M  and  N  any  two  integral  functions  or  constants.  Then 
every  common  factor  of  A  and  B  is  a  factor  of  MA  +  NB. 

For  let  F  denote  a  common  factor  of  A  and  B. 

Then  A^  GF       and        B=  HF, 

■where  G  and  H  are  integraL 

Hence    MA  +  NB  =  MGF+  NHF=  (MG  +  NH) F, 
where  ^^  +  Nil  is  integraL 

Therefore  i^  is  a  factor  of  MA  +  NB,  §  424. 

462  Applications  of  this  theorem.  By  aid  of  this  theorem  the 
problem  of  finding  the  H.C.F.  of  two  polynomials  in  x,  whose 
degrees  are  the  same,  may  be  reduced  to  that  of  factoring  a 
single  polynomial  of  a  lower  degree. 

Example  1.     Find  the  H.C.F,  of  ^  =  x2  +  2  x  -  4  and  B  =  x2  +  x  -  3, 
Subtracting  B  from  A,  we  obtain  A—  B  =  x  —  I. 
Hence,  §  461,  x  —  1  is  the  only  possible  common  factor  of  A  and  B. 
But,  since  A  does  not  vanish  when  x  =  1,  x  —  1  is  not  a  factor  of  A. 
Therefore  A  and  B  are  prime  to  one  another. 

Example  2.     Find  the  H.C.F.  of 

^  =  2x3-3x2-3x  +  2  and  JB  =  3x3  -  2x2  -  7x  -  2. 

1.    We  begin  by  multiplying  A  and  B  by  numbers  which  will  give 
results  having  the  same  leading  term,  namely,  ^  by  3  and  B  by  2. 
Then  subtracting  2  B  from  3  ^,  we  obtain 

3^-2J5  =  -5x2  +  5x+10=-5(x2-x-2)  =  -5(x+l)(x-2). 
Hence  the  only  possible  common  factors  of  A  and  Baiex  +  l  and  x  —  2. 


HIGHEST    COMMON    FACTOR  199 

By  the  remainder  theorem  we  find  that  both  are  factors  of  A  and  B. 
Hence  the  H.C.F.  of  ^  and  B  is  (a;  +  1)  (x  -  2). 
2.    Or  we  may  add  A  and  B,  thus  obtaining 

J.  +  B  =  5x3  -  5x2  -  lOx  =  5x(x2  -  X  -2)  =  5x(x  +  l)(x  -  2). 
It  is  at  once  evident  that  x  is  not  a  factor  of  A  or  B,  so  that  as  before 
we  have  only  to  test  x  +  1  and  x  -  2. 

Example  3.     Find  the  H.C.F.  of  the  following. 

1.  a;*  -  x3  +  3x2  -  4 X  -  12  and  x^  -  x^  +  2  x2  +  3 x  -  22. 

2.  6x3  +  25x2  +  5x  +  4  and  4x3  +  15x2 -2x  + 8. 

Theorem  3.     If  the  four  integral  functions  A,  B,  Q,  E,  are  so     463 
related  that  A  =  QB  +  E,,  the  common  factors  of  A  and  B  are 
the  same  as  the  common  factors  o/B  and  R. 

We  have  A=QB  +  R,  (1) 

and  therefore  A-  QB=  R.  (2) 

It  follows  from  (2),  by  §  461,  that  every  common  factor  of  A 
and  £  is  a  factor  of  R  and  therefore  a  common  factor  of  B  and  R. 

And,  conversely,  it  follows  from  (1),  by  §  461,  that  every 
common  factor  of  B  and  R  is  a.  factor  of  A,  and  therefore 
a  common  factor  of  A  and  B. 

Hence  the  common  factors  of  A  and  B  are  the  same  as  those 
of  B  and  R. 

The  general  method  for  finding  the  H.C.F.  of  two  polynomials     464 
in  X.     When  one  polynomial  in  x  is  divided  by  another,  the 
dividend,  divisor,  quotient,  and  remainder  are  connected  by 
the  identity  A  ^  QB  +  R.     Hence  it  follows  from  §  463  that 

The  common  factors  of  dividend  and  divisor  are  always  the 
same  as  those  of  divisor  and  remainder. 

By  making  use  of  this  fact,  the  H.C.F.  of  any  two  polyno- 
mials in  X  may  always  be  found.  The  method  is  analogous 
to  that  employed  in  arithmetic  to  find  the  greatest  common 
divisor  of  two  integers.  It  is  described  in  the  following  rule, 
where  yl  and  jS  represent  the  given  polynomials,  A  the  one  of 
higher  degree  if  their  degrees  are  not  the  same. 


200  A   COLLEGE   ALGEBRA 

465  Rule.  Divide  A.  hy  ^  and  call  the  quotient  q  and  the 
remainder  Rj. 

Next  divide  B  by  Rj  and  call  the  quotient  qi  and  the 
remainder  Rg- 

Next  divide  E-i  by  Rgj  «'^^  ■so  on,  continually  dividiiig  each 
new  remainder  by  the  one  last  obtained,  until  a  remainder  is 
reached  which  does  not  involve  x. 

If  this  final  remainder  is  not  0,  A  and  B  have  no  common 
factor.  If  it  is  0,  the  divisor  xvhich  yielded  it  is  the  H.C.F.  of 
A  and  B. 

For  suppose,  for  the  sake  of  definiteness,  that  the  final 
remainder  is  R3.  Then  according  as  (1)  R3  =  c,  where  c 
denotes  a  constant  7iot  0,  or  (2)  R^  =  0,  we  shall  have 

(1)     A    =qB     +R,       or        (2)     A   =qB     +  R, 

B    =  q,Ri  +  i?2  B    =  q,Ri  +  i2j 

i?i  =  q„R2  +  c  Ri=  q^Rz 

(1)  In  this  case  A  and  B  have  no  common  factor. 

For  it  follows  from  the  identities  (1),  by  §  463,  that  A  and 
B  have  the  same  common  factors  as  B  and  R^ ;  B  and  R^,  as  R^ 
and  i?2 ;  ^1  and  Ro,  as  R^  and  c. 

Hence  the  pairs  of  functions  A  and  B,  B  and  R^,  Ri  and  R2, 
Ro  and  c,  all  have  the  same  common  factors. 

But  as  c  is  a  constant  (not  0),  Ro  and  c  have  no  common 
factor.     Hence  A  and  B  have  none. 

(2)  In  this  case  Rg  *'«  ^^'«  H.C.F.  o/A  o7ic?  B. 

For  since  Ri  =  q^R^-,  every  factor  of  R^  is  a  common  factor 
of  Ri  and  R2,  and  R2,  itself  is  the  common  factor  of  highest 
degree. 

But  as  the  common  factors  of  R^  and  R^  are  the  same  as 
those  of  A  and  B,  the  factor  of  highest  degree  common  to  R^ 
and  ^2  is  also  the  factor  of  highest  degree  common  to  A  and  B. 
Hence  R.  is  the  H.C.F.  of  A  and  B. 


HIGHEST    COMMON    FACTOR  201 

Example  1.     Find  the  H.C.F.  of  x"  -\- x  +  1  and  x^  +  a;2  +  2  x  +  3. 
Writiug  divisors  at  tlie  left  of  dividends,  we  have 

£  =  x2  +  a;-fl|x3  +  a;-  +  2x  +  3[x  =  5 
x3  4-  x^  +     a: 
El  =  X  +  3|  X"  +     X  +  1  |x  —  2  =  gi 

x2  +  3  X 
-2x  +  1 


i?2=  7 

As  the  final  remainder,  i?2,  is  not  0,  x^  +  x  +  1  and  x^  +  x^  +  2  x  +  3 
have  no  common  factor. 

Example  2.     Find  the  H.C.F.  of 

x3  +  x2  +  2  X  +  2  and  x3  +  2  x2  +  3  X  +  2. 
Arranging  the  work  as  in  Ex.  1,  we  have 
B  =  x3  +  x2  +  2  X  +  2  I  x3  +  2  x2  +  3  X  +  2  1 1 
x3  -!■     x2  +  2  X  +  2 
Rx-  a;2+    X        |x3  +  x^  +  2x  +  2[x 

X3  +  X2 

E2  =  2x  +  2|x2  +  X  |x/2 

X2  +  X 
i?3=  0 

Here  the  division  by  E2  is  exact,  B.%  being  0.     Hence,  discarding  the 
numerical  factor  2  in  i?2)  we  have  H.C.F.  =  x  +  1. 

Here  for  the  first  time  we  have  an  actual  proof  —  for  f unc-  466 
tions  of  a  single  variable  • —  that  if  two  integral  functions 
have  any  common  factor,  they  have  a  highest  common  factor ; 
for  in  §§  463,  465  it  is  not  assumed  as  in  §  458  that  an  inte- 
gral function  can  be  expressed  in  but  one  way  in  terms  of  its 
prime  factors. 

Observe  that  in  the  proof  in  §  465  it  is  shown  that  467 

1.  Every  two  consecutive  fiinctioi^s  in  the  list  A,  B,  Ri.  R.,,  •  ■  - 
have  the  same  H.C.F.  as  A  and  B. 

2.  Every  common  factor  of  A  and  B  is  a  divisor  of  the  H.C.F. 
o/A  and  B. 

Abridgments  of  this  method.     1.    If  any  of  the  prime  factors     468 
of  ^  or  iJ  are  obvious  by  inspection,  begin  by  removing  these 


202  A   COLLEGE   ALGEBRA 

factors  and  then  find  the  H.C.F.  of  the  resulting  expressions. 
The  result  thus  obtained,  multiplied  by  such  of  the  factors 
removed  at  the  outset  as  are  common  to  A  and  B,  will  be  the 
H.C.F.  of  A  and  B,  §  458. 

The  same  course  may  be  followed  with  any  two  consecutive 
functions  in  the  list  A,  B,  R^,  R^,  ■■•,  since  every  two  such 
functions  have  the  same  H.C.F.  as  A  and  B,  §  467. 

Thus,  ^  =  a;<  +  a:3  +  2  x2  +  2  a;  and  B  =  X*  +  2  x3  +  3  x-^  +  2  a;  obviously 
have  the  common  factor  x.  Removing  it  we  have  x^  +  x-  +  2  x  +  2  and 
a;3  +  2x2  +  3x  +  2,  whose  H.C.F.  we  have  just  found  to  be  x  +  1  (see 
§465,  Ex.  2).     Hence  the  H.C.F.  of  A  and  B  is  x(x  +  1). 

Again  in  Ex.  2,  since  x  is  a  factor  of  i?i,  but  not  of  B,  it  cannot  be  a 
factor  of  the  H.C.F.  of  B  and  fii,  and  is  therefore  not  a  factor  of  the 
H.C.F.  of  A  and  B.  Hence  we  may  discard  this  factor  x  of  Ei  and  divide 
B  by  the  remaining  factor  x  +  1,  so  lessening  the  number  of  divisions. 

2.  In  any  of  the  divisions  we  may  multiply  or  divide  the 
divisor  or  dividend  or  any  intermediate  remainder  by  a  numer- 
ical factor ;  for  this  will  affect  the  subsequent  remainders  by 
a  numerical  factor  at  most,  §  403,  and  therefore  the  H.C.F.  not 
at  all.  This  device  enables  us  to  avoid  fractional  coefficients 
when  the  given  coefficients  are  rational. 

3.  It  is  advantageous  to  employ  detached  coefficients. 

Example  1.     Find  the  H.C.F.  of 

^  =  X*  +  .3x3  +  2x2  +  3x  +  1  and  B  =  2x^  +  5x2  -  x  -  1, 
Multiplying  Ahy  2  and  using  detached  coefficients,  we  have 


2+5-1-1 

|2  +  6+    4  + 
2  +  5-    1  - 

1+    5  + 

2 

2  +  10  + 

2+    5- 

6  +  2[l 

1 
7+2 

14  +  41_1_ 
1  -1 

Hence  the  H.C.F, 

isx2  +  3a 

5)5  +  15  +  5 

1+    3  +  l|2  +  5- 

2  +  6  + 

,  of  ^  and  B                     -  1  - 

;  +  l.                                   -1- 

2 

3- 

3- 

-1| 
-J 

HIGHEST   COMMON    FACTOR 


203 


This  reckoning  may  be  more  compactly  arranged,  as  follows ; 

1  +  1 


2+5-1-1 
2  +  6  +  2 

2  +  6+    4+    6  +  2 
2  +  5-    1  -    1 

-1-3-1 
-1-3-1 

1+    5+    7  +  2 
2  +  10  +  14  +  4 
2+    5  -    1-1 

5  +  15  +  5 
1+3  +  1 

Example  2.     Find  the  H.C.F.  of  the  following. 

2  X*  +  3  x3  +  4  a;2  +  2  X  +  1  and  2  x*  -  x^  +  2  x2  +  i. 

Theorem.     If  the  coefficients  of  A  and  B  are  7'ational,  so  are    469 
those  of  their  H.C.F. ;   and  if  the  coefficients  of  A  and  B  are 
real,  so  are  those  of  their  H.C.F. 

For  the  H.C.F.  of  A  and  B  can  be  found  by  the  method  of 
§  465.  Hence  its  coefficients  are  rational  combinations  of  the 
coefficients  of  A  and  B,  and  are  therefore  rational  when  these 
are  rational,  real  when  these  are  real. 

This  theorem  has  important  consequences,  some  of  which 
will  be  noticed  later.  By  applying  it,  we  can  often  shorten 
the  work  of  finding  a  H.C.F. 

Example.     Find  the  H.C.F.  of  x2  -  2  and  x^  +  x^  -  5x  +  6. 

Either  these  polynomials  are  prime  to  one  another,  or  their  H.C.F.  is 
x'-^  —  2  itself  ;  for  since  the  factors  of  x^  —  2,  namely  x  +  V2  and  x  —  V2, 
have  irrational  coefficients,  neither  of  them  can  be  the  H.C.F. 

But  by  trial  we  find  that  x^  +  x^  —  5  x  +  6  is  not  exactly  divisible  by 
x2  —  2.     Hence  these  polynomials  are  prime  to  one  another. 

The  H.C.F.  of  more  than  two  polynomials  in  x.     This  may  be     470 
obtained  by  the  following  rule. 

First  find  the  H.C.F.  of  two  of  the  polynomials,  next  the 
H.C.F.  of  the  result  and  the  third  polynomial,  and  so  on.  The 
final  result  ivill  be  the  H.C.F.  required. 

Thus,  if  D^  is  the  H.C.F.  of  A  and  B,  and  D^  is  the  H.C.F. 
of  Di  and  C,  then  D^  is  the  H.C.F.  oi  A,  B,  and  C. 


204  A   COLLEGE    ALGEBRA 

For  (1),  D2  is  a  factor  of  D^  and  C ;  and  Dy  is  a  factor  of 
A  and  B.     Hence,  §  427,  D^  is  a  factor  of  ^,  B,  and  C. 

And  (2),  every  common  factor  of  A,  B,  and  C  is  a  common 
factor  of  Di  and  C,  and  therefore  a  factor  of  D2,  §  467.  Hence 
D2  is  the  highest  common  factor  of  A,  B,  and  C. 

The  same  conclusion  follows  from  §  458. 

Example.     Find  the  H.C.F.  oi  A  =  x*  +  x^  -  x"  +  x  -  2, 

B  =  2  X*  +  5x3  -  2  x2  -  7  X  +  2,  and  C  =  3x*  -  x^  -  x^  -  2. 

By  §  465,  we  find  that  the  H.C.F.  of  ^  and  J5  is  A  =  a;2  +  x  -  2  ;  and 
that  the  H.C.F.  of  Di  and  C  is  x  -  1. 

Hence  the  H.C.F.  of  A,  B,  and  C  is  x  -  L 

471  The  H.C.F.  of  polynomials  in  more  than  one  variable.  The 
general  problem  of  finding  the  H.C.F.  of  two  such  polynomials 
is  too  complicated  to  be  considered  here.  But  the  H.C.F.  of 
two  polynomials  which  are  homogeneous  functions  of  tioo  vari- 
ables, as  X  and  y,  may  readily  be  found  by  aid  of  the  rule 
given  in  §  465. 

EXERCISE   XXn 
Find  the  H.C.F.  of  the  following. 

1.  10  xhf-z^,  4  x^yz^,  6  x^yH^,  and  8  x^y^z^u. 

2.  {a  +  6)2  (a  -  6),  (a  +  b)  (a  -  6)2,  and  a^b  -  ab^ 

3.  2/*  +  2/2  +  1  and  y'^  -  y -\.  I. 

4.  a2  -  1,  a2  +  2  a  +  1,  and  a^  +  1. 

5.  x'  -  1  and  x^  +  ax2  -  ax  -  1. 

6.  X*  —  y*,  x^  +  2/",  and  x^  +  x'hj  +  xy'^  +  y^. 

7.  x2  +  5  X  +  6,  x2  +  X  -  2,  and  x-  -Ux-  32. 

8.  (x-l){x-2)  and5x<-15x3  +  8x2  +  6x-4. 

9.  x'  —  1  and  x^  —  4  x2  —  4  x  —  5. 

10.  (x2  -  1)2  (X  +  1)2  and  (x-^  +  5  x2  +  7  x  +  3)  (x2  -  6  x  -  7). 

11.  (X  -  1)2  (X  -  2)2  and  (x2  _  3  x  +  2)  (2  x^  -  5  x2  +  5  x  -  6). 

12.  2  x3  -  3  x2  -  11  X  +  6  and  4  x3  +  3  x2  -  9  X  +  2. 

13.  x8  -  2  x2  -  2  X  -  3  and  2  x8  +  x2  +  X  -  1. 


HIGHEST   COMMON   FACTOR  205 

14.  3  z3  +  2  x2  -  19  a;  +  6  and  2  x3  +  x2  _  13  a;  ^  6; 

15.  X*  -  x3  -  3  x2  +  a;  +  2  and  2  X*  +  3  x3  -  x2  -  3  X  -  1. 

16.  3  x3  -  13  x2  +  23  X  -  21  and  6  x^  +  x^  -  44  x  +  21. 

17.  3  x3  +  8  x2  -  4  X  -  15  and  6  X*  +  10  x3  -  3  x2  -  2  X  +  5. 

18.  G  x5  4-  7  a;4  _  9  x3  -  7  x2  +  3  X  and  G  x^  +  7  x*  +  3  x^  +  7  x2  -  3  x. 

19.  6  x<  -  3  x3  +  7  x2  +  x  -  3  and  2  X*  +  3  x3  +  7  x2  +  3  X  +  9. 

20.  6x5-4x*-llx3-3x2-3x-l  aud4x*  +  2x3-  18x2 +  3x- 5. 

21.  x5  -  x3  -  4  x2  -  3  X  -  2  and  5  x*  -  3  x2  -  8  x  -  3. 

22.  3  x3  _  x2  _  12  X  +  4,  x3  -  2  x2  -  5  X  +  6,  and  7  x^  +  19  x2  +  8  x  -  4. 

23.  x3  +  ax2  -  3  X  -  3  a,  x3  -  x2  -  3  X  +  3,  and  x^  +  x2  -  3  x  -  3. 

24.  7  x%  -  G  xhf-  - 1 8  xh/^  +  4  xy*  and  14  xSy  - 19  x'^y^  -  32  xy^  +  28  y*. 

25.  X  (X  -  1)  (x3  +  4  x2  +  4  X  +  3)  and  (x  -  1)  (x  +  3)  (12  x3  +  x2  +  x  - 1), 

26.  4  x3  -  8  x2  -  3  X  +  9  and  (2  x2  -  x  -  3)  (2  x2  -  7  x  +  6). 

LOWEST   COMMON   MULTIPLE 

Lowest  common  multiple.     A  common  multiple  of  two  or  more     472 
integral  functions,  A,  B,---,  is  an  integral  function  which  is 
exactly  divisible  by  each  of  the  functions  A,  B,--. 

Among  such  common  multiples  there  is  one  whose  degree  is 
lowest.  We  call  this  the  loivest  common  multiple  (L.C.M.)  of 
A,  B,--. 

Theorem  1.      The  L.C.M.  of  tivo  or  more  integral  functions,     473 
A,  B,  •  •• ,  is  the  product  of  all  the  different  prime  factors  of 
A,  B,  •  ■  •,  each  raised  to  the  highest  power  in  which  it  occurs  in 
any  of  these  functions. 

This  follows  from  the  fact  that  a  common  multiple  oi  A,B,-  • 
must  contain  every  prime  factor  of  each  function  A,  B,---  dX 
least  as  often  as  it  occurs  in  that  function,  hence  all  the 
factors  mentioned  in  the  theorem.  And  the  common  multiple 
of  lowest  degree,  that  is,  the  L.C.M.,  is  the  one  which  contains 
no  factors  besides  these. 


206  A   COLLEGE   ALGEBRA 

Here,  as  previously,  we  ignore  numerical  factors  and  assume  that  an 
integral  function  can  be  expressed  in  only  one  way  as  a  product  of  powers 
of  its  different  prime  factors. 

474  Finding  the  L.C.M.  by  inspection.  If  we  can  resolve  A,  B,  ■■■ 
into  their  prime  factors,  we  may  obtain  their  L.C.M.  at  once 
by  applying  the  theorem  just  demonstrated. 

Example  1.     Find  the  L.C.M.  of  Sx^y'^z,  xy*z^,  and  2x-yz^. 
Here  the  different  prime  factors,  each  raised  to  the  highest  power  iu 
which  it  occurs  in  any  of  the  functions,  are  x^,  ?/*,  z^. 
Hence  the  L.C.M.  is  x^yH^. 

Example  2.     Find  the  L.  C.  M.  of  xV  _  4  x?/2  +  4  ?/2  and  x"y  -  4  ?/. 
We  have  x2?/2  -  4  xy2  +  4  2/2  =^2  (a;  _  2)2  and  x22/  -  4  2/  =  2/(x  -  2)(x  +  2). 
Hence  the  L.C.M.  is  2/2(3;  _  2)2 (x  +  2). 

475  Theorem  2.  The  L.C.M.  of  two  integral  functions,  A  and  B, 
is  the  jjroduct  of  the  two,  divided  by  their  H.G.F. 

For  let  D  denote  the  H.C.F.  of  A  and  B,  and  let  A^  and  B^ 
denote  the  quotients  obtained  by  dividing  A  and  B  by  D,  so  that 
A  =  AiD  and  B  =  B^D. 

Then  if  M  denote  the  L.C.M.  of  A  and  B,  we  have 
M=  A^B^D=  AB/D. 

For  evidently  a  common  multiple  of  A  and  B  must  contain 
(1)  the  product  of  all  prime  factors  common  to  A  and  B, 
namely  D,  (2)  the  product  of  all  prime  factors  of  A  not  belong- 
ing to  B,  namely  Ai,  (3)  the  product  of  all  prime  factors  of 
B  not  belonging  to  A,  namely  B^;  and  the  loivest  common 
multiple  will  contain  no  factors  besides  these. 

476  Corollary.  The  product  of  two  integral  f  motions,  A  and  B, 
is  equal  to  the  product  of  their  L.C.M.  ayid  their  H.C.F. 

477  General  method  for  finding  the  L.C.M.  of  two  polynomials  in  x. 

It  follows  from  §§  4G5,  475  that  the  L.C.M.  of  two  such  poly- 
nomials, A  and  B,  may  always  be  obtained  by  the  rule : 

To  find  the  L.C.M.  of  A  and  B,  divide  A  by  the  H.C.F.  of 
A  and  B,  and  multiply  the  result  by  B. 


HIGHEST   COMMON    FACTOR  207 

Observe  that  this  is  equivalent  to  multiplying  B  by  all  those 
prime  factors  of  A  which  are  not  already  present  in  B. 

Example.     Find  the  L.C.M.  of 

X*  +  3  x3  +  2  X-  +  3  X  +  1  and  2  x-'^  +  5  x^  -  x  -  1. 

By  §  465,  we  find  that  the  H.C.F.  =  x'-  +  3x  +  1. 

Again  (2x3  +  5x2  -x  -  l)/(x2  +  3x  +  1)  =  2x  -  1. 

Hence  the  L.C.M.  is      (x*  +  3 x-^  +  2 x'-  +  3  x  +  1)  (2  x  -  1). 

The  L.C.M.  of  more  than  two  polynomials  in  x.     This  may  be     478 
obtained  by  the  following  rule. 

First  find  the  L.C.M.  of  two  of  the  polynomials,  next  the 
L.C.M.  of  the  result  and  the  third  polynomial,  and  so  on.  The 
final  result  will  be  the  L.C.M.  required. 

This  follows  from  the  fact  that  each  step  in  the  process  is 
equivalent  to  multiplying  the  L.C.M.  last  obtained  by  those 
prime  factors  of  the  next  function  which  are  not  already 
present  in  that  L.C.M. 

Example.     Find  the  L.C.M.  of  J.  =  x*  +  3  x^  +  2  x'^  +  3  x  +  1, 

B  =  2  x^  +  5  x2  -  X  -  1,  and  O  =  2  x3  -  3  x2  +  2  x  -  3. 
As  we  have  just  shown,  §  477,  Ex.,  the  L.C.M.  of  A  and  B  is 

Jlfi  =  (X*  +  3  x3  +  2  x2  +  3  X  +  1)  (2  X  -  1). 
We  have  next  to  find  the  L.C.M.  of  Mi  and  C. 

By  division  we  find  that  2  x  —  1  is  prime  to  C,  and  by  §  465  we  find 
that  the  H.C.F.  of  x*  +  3  x^  +  2  x^  +  3  x  +  1  and  C  is  x2  +  1. 
Furthermore,  C/  (x-  +  1)  =  2  x  -  3. 

Hence  the  L.C.M.  of  Mi  and  C,  and  tlierefore  of  A,  B,  C,  is 

ilf  =  (X*  +  3  x3  +  2  x2  +  3  X  +  1)  (2  X  -  1)  (2  X  -  3). 
Observe  that  we  do  not  multiply  the  factors  of  Mi  together  before 
proceeding  to  find  the  H.C.F.  of  Mi  and  C. 

EXERCISE  XXra 
Find  the  L.C.M.  of  the  following. 

1.  3x  -  1,  9x2  _  1^  and  9x2  +  1. 

2.  (a  +  6)  (a5  -  tfi)  and  (a  -  b)  (a^  +  66). 


208  A    COLLEGE    ALGEBRA 

3.  a^  +  a2  _|_  a,  a^  —  a^,  and  a^  —  a^. 

4.  (x3  _  y3)  (X  _  2/)3,  (x4  _  ^Z*)  (x  -  yf,  and  (x2  -  2/2)2. 

5.  x2  -  3  X  +  2,  x2  -  5  X  +  6,  and  x2  -  4  X  +  3. 

6.  x2  -  (1/  +  z)2,  2/2  _  (2  ^  x)2,  and  22  _  (^  +  y)2. 

7.  2  x2  +  3  x?/  -  9  ?/2,  3  x2  +  8  x?/  -  3  ?/2,  and  6  x2  -  11  xy  +  3  y2. 

8.  x3  +  x2  +  X  +  1  and  x''  -  xr  +  x  -  1. 

9.  2  a2x  +  2  x^y  +  3  y^x  +  3  a^-y  and  (2  x2  -  3  a2)  ?/  +  (2  a2  -  3  2/2)  x. 

10.  8  x3  -  18  x?/2,  8  x3  +  8  x2y  -  6  xy",  and  8  x2  -  2  xy  -  15  2/2. 

11.  x3  +  2/3,  x3  —  2/3,  and  x*  +  x22/2  +  y*. 

12.  x6  -  1,  3  x3  -  5  x2  -  3  X  4  5,  and  x*  -  1. 

13.  8x3  +  27,  16x*  +  36x2  +  81,  and  6x2  +  5^-6. 

14.  x2  -  4  a\  x3  +  2  ax2  +  4  a2x  +  8  a^,  and  x3  -  2  ax2  +  4  a2x  -  8  a^. 

15.  x2  +  2  X,  x2  +  6x  +  2  X  +  2  6,  and  x3  +  ax2  -  b-x  -  aW-. 

16.  (x2  +  3  X  +  2)  (x2  +  7  X  +  12)  and  (x2  +  5  X  +  6)  (2  x2  -  3  X  -  5). 

17.  (x3  -  8)  (27  x3  +  1)  and  (2  x3  +  5  x2  +  10  x  +  4)  (x3  -  x2  -  x  -  2). 

18.  x3-6x2  +  llx-6,  2x3-7x2  +  7x-2,  and  2x3  +  x2  -  13x  +  6. 

19.  x*  +  5  x2  +  4  X  +  5,  2  x'*  -  x3  +  10  x2  +  4  X  +  5,  and 

2x*  +  x3  +  7x2  +  3x  +  3. 

20.  2x*  -  x3  +  2x2  +  3x  -  2,  2x*  +  3x3  -  4x2  +  is^  _  q^  and 

X*  +  3  x3  +  x2  +  5  X  +  6. 

ON  THE  PRIME  AND   IRREDUCIBLE  FACTORS   OF  FUNCTIONS 
OF   A   SINGLE   VARIABLE 

In  the  following  theorems  A  and  B  denote  polynomials  in  x. 

479         Fundamental  Theorem.     //"A  is  prime  to  B,  tivo  integral  func- 
tions, M  and  N,  can  he  found  such  that 
MA  +  NB  =  1. 

For  if  we  apply  the  method  of  §  465  to  A  and  B,  we  shall  obtain  as 
final  remainder  a  constant,  c,  different  from  0. 


HIGHi:.ST   COMMON    FACTOR  209 

If  Tve  suppose,  as  in  §  405,  that  c  is  the  third  remainder,  and  use  the 
notation  there  explained,  we  have 

1.  A  =qB     +  Ri,  and  therefore   4.    J[?i  =  J    -  qB, 

2.  B  =5iEi  +  i?2,  5.    E2  =  B  -  qiEi, 

3.  Ei  =  q2li2  +  C,  6.     c      =i?i-goi?2. 

Substitute  in  6  the  value  of  i?2  given  by  5,  collecting  the  i?i  and 
the  B  terms,  and  in  the  result  substitute  the  value  of  Ri  given  by  4, 
collecting  the  A  and  B  terms.     We  thus  obtain 

c  =  RiqnR2 
=  {l-\-qiq'z)Ri-q,B 
=  (1  +  QlQi)  A-(q  +  q.  +  qqiqo)  B. 

Divide  both  sides  of  this  last  identity  by  c,  and  for  (1  +  qiq^) /c  and 
—  (7  +  92  +  QQil^l/c,  which  are  integral  functions  since  c  is  a  constant, 
write  M  and  N.     We  obtain 

1  =  MA  +  NB, 

where,  as  just  said,  M  and  N  are  integral  functions. 

And  we  may  demonstrate  the  theorem  in  the  same  way  when  the  con- 
stant remainder,  c,  is  obtained  earlier  or  later  than  the  third  division. 

Conversely,  If  MA  +  NB  =  1,  where  M  and  N  are  integral,     480 
then  A  is  jjrime  to  B. 

For  a  common  factor  of  A  and  B  would  be  a  factor  of  MA  +  NB^ 
§  461,  and  therefore  of  1,  which  is  impossible. 

The  following  theorems  are  some  of  the  more  important 
consequences  of  the  fundamental  theorem  just  demonstrated. 

Theorem  1.     7/*  A  is  prime   to   B,   and   the  j^^'oduct   AC   is     481 
divisible  by  B,  then  C  is  divisible  by  B. 

For  since  A  is  prime  to  B,  we  can  find  M  and  iV,  §  479,  such  that 
MA  +  NB  =  -l, 
and  therefore  M-AC+NCB=C. 

But  B  is  a  factor  of  both  A  C  and  B.     Hence  it  is  a  factor  of  C,  §  461. 

Theorem  2.     Tf  A  is  jjrime  to  each  of  the  functions  B  and  C,     482 
it  is  prime  to  their  product,  BC. 


210  A   COLLEGE   ALGEBRA 

For  since  A  Is  prime  to  B,  we  can  find  M  and  N,  §  479,  such  that 
MA  +  NB=1, 
and  therefore  MC  ■A  +  NBC=C. 

Hence,  if  A  and  BC  had  a  common  factor,  it  would  be  contained  in  C, 
§  46L     But  this  is  impossible,  since  A  is  prime  to  C. 

483  Corollary.     If  A  is  j^rlme  to  each  of  the  functions,  B,  C,  D, 
and  so  on,  it  is  2Ji'iine  to  their  2}>'od>(ct,  B  ■  C  •  D  •  •  •. 

For,  as  just  demonstrated,  A  is  prime  to  BC. 

And  since  A  is  prime  to  BC  and  also  to  D,  it  is  prime  to  the  product 
BCD  ;  and  so  on. 

484  Theorem  3.     A  composite  function  has  one  and  but  one  set  of 
jjrime  factors. 

For  let  P  denote  the  given  function  and  n  its  degree. 

If  P  is  composite  it  has  some  factor  A.  If  ^4.  in  turn,  is  composite,  it 
has  some  factor  B.  Continuing  thus,  we  must  ultimately  come  upon  a 
prime  function  ;  for  the  degrees  of  the  successive  functiims  P,  A,  B,  ■■• 
begin  with  the  finite  number  n,  decrease,  and  cannot  fall  below  1. 

Let  F  denote  this  prime  function.  It  is  one  of  the  prime  factors  of  P, 
§  427,  and  we  have  P=  FM,  where  M  is  integral. 

Similarly  if  M  is  composite,  a  prime  function  F'  exists  such  that 
M=  F'M',  and  therefore  P  =  FF'M',  where  M'  is  integral. 

Continuing  thus,  we  reach  the  conclusion  that  a  .series  of  prime  func- 
tions F,  P',  P",  ■  •  ■  exists,  whose  number  cannot  exceed  n,  such  that 
P  =  F-F'F"  ■■■. 

Hence  P  has  at  least  one  set  of  prime  factors. 

Moreover  P  can  have  but  one  such  set  of  factors.     For,  suppose  that 

P  =  FF'F"--  =  G-G'-  G"  ■■■ 

where  G,  G',  G",  •  •  ■  also  denote  prime  functions. 

Then  G  cannot  be  prime  to  all  the  functions  P,  P',  P",  •  •  ■ ,  for,  if 
so,  it  would  be  prime  to  their  product  P,  §  483,  whereas  it  is  a  factor 
of  P. 

Suppo.se,  therefore,  that  G  is  not  prime  to  P,  for  example.  Then  G 
and  P  have  a  common  factor.  But  G  and  P  are  prime  functions,  and 
two  prime  functions  can  have  no  factor  in  x  but  themselves  in  common. 
Hence  G  differs  from  P  by  a  numerical  factor,  as  c,  at  most,  and  we  have 
G  =  cF. 


HIGHEST   COMMON   FACTOR  211 

But  substituting  this  value  of  G  in  the  identity  FF'F"  ••  ■  =  GG'G"  •  •  •, 
and  dividing  both  members  by  F,  we  have 

F'F"---^cG'G"---, 

from  which  it  follows  by  a  mere  repetition  of  our  reasoning  that  G'  differs 
from  one  of  the  functions  F\  F",  •  •  •  by  a  numerical  factor  at  most. 

Continuing  thus,  we  reach  the  conclusion  that  the  set  of  fvmctions  G, 
G\  G",  •  •  •  differs  from  the  set  F,  F',  F",  •  ■  •  at  most  by  numerical  factors 
or  in  the  order  in  which  they  are  arranged. 

Corollary.     A  composite  function  can  be  expressed  in  only  one     485 
way  as  a  product  of  powers  of  its  different  prime  factors. 

This  follows  at  once  from  the  identity  P  =  i^-  F'  •  F"  •  •  • ,  if  we  replace 
each  set  of  equal  factors  in  the  product  F  ■  F'  ■  F"  ■  ■  •  by  the  correspond- 
ing power  of  one  of  these  factors. 

Irreducible  factors.     By  the  irreducible  factors  of  an  integral    486 
function  with  rational  coefficients,  we  usually  mean  the  factors 
of  lowest  degree  with  rational  coefficients. 

Thus,  while  the  prime  factors  of  (x  —  1)  (x^  —  2)  are  x  —  1,  x  —  v^, 
X  +  V2,  the  irreducible  factors  are  x  —  1  and  x^  —  2. 

From  the  theorems  just  demonstrated  and  the  theorem  of    487 
§  469,  it  follows  that 

A  reducible  integral  function  with  rational  coefficients  can  be 
expressed  in  only  one  way  as  a  product  of  powers  of  its  different 
irreducible  factors. 


DIGRESSION   IN   THE  THEORY  OF  NUMBERS 

Theorems  analogous  to  those  just  demonstrated  hold  good    488 
for  integral  numbers. 

We  shall  employ  the  letters  a,  b,  and  so  on,  to  represent 
integers,  positive  or  negative  (not  0),  and  shall  mean  by  a 
factor  of  a  any  integer  which "  exactly  divides  a. 

A  prime  number  is  an  integer  which  has  no  other  factors     489 
than  itself  and  1. 


116|325[2 

232 

ri=   93|ll6[l 

93 

r2=  23|tt3 

[i 

92 

n=  1 

Hence,  starting 

w 

212  A    COLLEGE    ALGEBRA 

490  If  two  integers,  a  and  b,  have  no  common  factor  except  1, 
a  is  said  to  be  prime  to  b. 

491  Theorem,     i/'a  is  prime  to  b,  two  integers,  m  and  n,  can  always 

be  found  such  that 

ma  +  nb  =  1. 

For  since  a  is  prime  to  6,  if  we  apply  the  usual  method  for  finding  the 
greatest  common  divisor,  we  shall  obtain  1  as  the  final  remainder.  We 
may  deduce  the  theorem  from  this  fact  by  the  reasoning  of  §  479. 

Thus,  let  a  -  325,  b  =  116.  Applying  the  method  for  finding  G.C.D., 
we  have 


325  =  2-110  +  93,  or  93  =  325-2- 116     (1) 

116  =  1  -    93  +  23,  or  23  =  116  -  1  •    93     (2) 

93  =  4  •    23  +    1,  or     1  =    93  -  4  -    23     (3) 
fith  (3),  and  substituting  first  the  value  of  23  given 
by  (2),  and  then  the  value  of  93  given  by  (1),  we  have 
1  =  93  -  4  .  23 
=.5.93 -4.116 
=  5-325-  14-  116. 
Therefore  5  .  325  +  (- 14)  - 116  =  1. 

Hence  we  have  found  two  integers,  m  =  5  and  ?i  =  —  14,  such  that 

?u-325  +  n-  116  =  1. 
And  similarly  in  every  case. 

Example.     Find  integers  m  and  ?i  such  that  223  m  +  125  n  =  1. 
492         Corollaries.     From  this  fundamental  theorem  we  may  derive 
for  integral  numbers  theorems  analogous  to  those  derived  for 
integral  functions  in  §§  481-485,  and  by  the  same  reasoning. 
In  particular  we  may  prove  that 

1.  If  Si  is  ])ri?ne  to  b,  a/id  the  product  ac  is  divisible  by  b, 
then  c  is  divisible  by  b. 

2.  If  a,  is  prime  to  b  and  c,  then  a  is  prime  to  be. 

3.  A  composite  number  can  be  expressed  in  one  ivay,  and  hut 
one,  as  a  product  of  powers  of  its  different  prime  factors. 


RATIONAL   FRACTIONS  213 

VIII.     RATIONAL   FRACTIONS 

REDUCTION   OF   FRACTIONS 

Fractions.     Let  A  and  B  denote  any  two  algebraic  expres-    493 
sions,  of  which  B  is  not  0.     The  quotient  of  A  by  B,  expressed 
in  the  form  A  / B,  is  called  a  fraction ;  and  A  is  called  the 
numerator,  B  the  denominator,  and  .1  and  B  together  the  terms 
of  this  fraction. 

When  both  A  and  B  are  rational,  A  ]  B  is  called  a  rational    494 
fraction. 

When  both  A   and  B  are  integral,  A  / B  \s  called  a  simple     495 
fraction ;  but  if  .4  or  B  is  fractional,  A  /  B  is  called  a  complex 
fraction. 

A  simple  fraction  is  called  a  proper  or  an  improper  fraction,     496 

according  as  the  degree  of  its  numerator  is  or  is  not  less  than 

that  of  its  denominator. 

^,         x-y        ,2z2_3  2a;2+i  x^  -  3  . 

Thus,  and  are  proper, and  improper, 

X2  +  2/2  x3  +  1  ^      '  X2  +  1  x2  +  1 

An  improper  fraction  whose  terms  are  functions  of  a  single    497 
variable  can  be  reduced  to  the  sum  of  an  integral  expression 
and  a  proper  fraction,  §  400.     This   sum   is  called  a  mixed 
expression. 

T,,                      2x2+1                  1             x3-3              x  +  3 
Thus,  =  2  —  - ,       =  x . 

X2  +  1  X2  +  1  X2  +  1  X2  +  1 

Allowable  changes  in  the  form  of  a  fraction.     These  depend  on     498 
the  following  theorem,  §  320,  1. 

The  value  of  a  fraction  remaij^s  unchanged  ichen  its  mimer- 
ator  and  denominator  are  multiplied  or  divided  by  the  same 
expression  (not  0). 

In  particular,  we  inay  change  the  signs  of  both  numerator  aiid  denomina- 
tor, this  being  equivalent  to  multiplying  both  numerator  and  denominator 
by  —  1.  Changing  the  sign  of  the  numerator  or  of  the  denominator  alone 
will  change  the  sign  of  the  fraction  itself,  §  320,  3. 


214  A    COLLEGE   ALGEBRA 

If  the  numerator  or  denominator  be  a  polynomial,  changing  its  sign  is 
equivalent  to  changing  the  signs  of  all  its  terms. 

a+h—c      c—a—b  c—a—b  a+b—c 

Thus, =  7 = r =  -7 

a  —  b  +  c      b  —  c  —  a         a  —  b  -j-  c  b  —  c  —  a 

K  the  numerator,  denominator,  or  both,  are  products  of  certain  factors, 
we  may  change  the  signs  of  an  even  number  of  these  factors ;  but  chang- 
ing the  signs  of  an  odd  number  of  them  will  change  the  sign  of  the 
fraction. 

Thu       (a  -b){c-  d)  ^  (b  -  a)  {c  -  d)  ^  _  (b  -a){d-  c) 
''^'     (e-f){g-h)      (e-f){h-g)  (f-e){g-h)' 

499  Reduction  of  fractions.  To  simjdifi/  a  fraction  is  to  cancel  all 
factors  which  are  common  to  its  numerator  and  denominator, 
this  being  a  change  in  the  form  of  the  fraction  which  will  not 
affect  its  value,  §  498. 

When  this  has  been  done  the  fraction  is  said  to  be  in  its 
lowest  terms,  or  to  be  iri'educihle. 

We  discover  what  these  common  factors  are,  or  show  that 
there  are  none,  by  the  methods  of  Chapter  VII.  W"e  look 
first  for  common  monomial  factors  and  other  common  factors 
which  are  obvious  by  inspection  or  which  can  be  foimd  by  aid 
of  the  remainder  theorem,  and  when  these  simpler  methods 
fail  we  apply  the  general  method  of  §  465. 

The  following  examples  will  illustrate  some  of  these  methods. 

Example  1.     Simplify  (aec  —  ade)  /  {bde  —  ebc). 

„„    ,  aec  —  ade      ae  (c  —  d)  a(c  —  d)  a 

We  have =  — ^^ '- = ^ -  = 

bde  —  ebc      be  {d  —  c)  b{c  —  d)  b 

Example  2.     Simplify  {x^  +  x"  +  x  +  0)/{x-  +  3  x  +  2). 

By  inspection,  the  factors  of  the  denominator  are  x  +  1  and  x  +  2. 
Hence  if  numerator  and  denominator  have  any  common  factor,  it  must 
be  one  of  these.  Testing  by  synthetic  division,  we  find  that  the  numer- 
ator is  not  divisible  by  x  +  1,  but  is  divisible  by  x  +  2,  the  quotient  being 
x2  -  X  +  3, 

x^  +  x^  +  x  +  6      x2-x-f-3 


Hence 


x2-l-3x  +  2  x  +  1 


RATIONAL    FRACTIONS  215 

Example  3.     Simplify  (x^  +  7  x  +  10)  /  (x^  +  5  x  +  6). 
Subtracting  denominator  from  numerator,  we  have 

x3  +  7  X  +  10  -  (x3  +  5  X  +  G)  =  2  (X  +  2). 

Hence,  if  the  numerator  and  denominator  have  any  common  factor,  it 
must  be  X  +  2,  §  461.  But  the  numerator  does  not  vanish  when  x  =  —  2. 
Hence,  §  415,  the  fraction  is  already  in  its  lowest  terms. 

Example  4.     Simplify  — ^^ '—^ ^^ '— !^ '-. 

{a  -b)(b-  c)  (c  -  a) 

Here  the  only  possible  common  factors  are  a  —  b,  b  —  c,  and  c  —  a. 
Setting  a  =  6  in  the  numerator,  we  have  b^(b  ~  c)  +  b-{c  —  b),  or  0. 
Hence,  §  417,  the  numerator  is  divisible  by  a  —  6.  And  we  may  show 
in  the  same  way  that  it  is  divisible  by  6  —  c  and  c  —  a. 

Therefore  the  numerator  is  exactly  divisible  by  the  denominator.  But 
the  two  are  of  the  same  degree,  namely  three,  in  a,  b,  c.  Their  quotient 
must  therefore  be  a  mere  number ;  and  since  the  a^  terms  in  the  two, 
when  arranged  as  polynomials  in  a,  are  a-  (6  —  c)  and  —  a^  (6  -  c)  respec- 
tively, this  number  is  —  1. 

Hence  the  given  fraction  is  equal  to  —  1. 

Example  5.     Simplify  (2x3  + 13x2-0x  +  7)/{2x*  +  5x-5  +  8x2-2x  +  5). 

By  §465,  we  find  that  the  H.C.F.  of  numerator  and  denominator 
is  2x2  — x  +  1.  And  dividing  both  , numerator  and  denominator  by 
2  x2  _  X  +  1,  we  obtain 

2x3  +  13x2-6x  +  7  x  +  7 


2x4  +  5x3  + 8x2- 2x  + 5      x'^  +  3x  +  5 

EXERCISE  XXIV 
Reduce  the  following  fractions  to  their  lowest  terms. 
x5y3  _  4  x^y^  _      (x6  -  y^)  (X  +  y) 


X^y^  -  2  X^^  (X3  +  y3)  (x4  _  ^4) 

x2_4x-21  .     3x2 -8x- 3 


x2  +  2x-63  3x2  +  7x  +  2 

3x2-  186x  +  27b2  5x2 +  6ax  + a' 

2x2-1862        *  ■    5x2  +  2ax-3a2' 

(x2-25)(x2-8x  +  15)  g     15x2-46x  +  35 

{x2-9)(x2-7x  +  10)  '  ■    10x2-29x  +  21* 


216  A   COLLEGE   ALGEBRA 


a;*4-a;V  +  1/*  -^q    x^  -  tj^  +  z"^  +  2  xz 


(a;3  4.  ^3)  (a;3  _  ^^3)  x:^  ^  yz  _  z''^  +  2  xy 

(1  +  xyY  -  (X  +  vY  ,2     2»ix  -  my  -  12?ix  +  6ny 

1  —  x2  '    6  mx  —  3  »i2/  —  2  Tix  +  nj/ 

2 x3  +  7 x^  -  7 X  -  12  x3-8x2  + 19X-12 


15. 


2x3  +  3x2-14x- 15  2x3-13x2  +  17x4-12 

x*  +  x3  +  5x2  +  4x  +  4  ,„  x3-2x2-x-6 


2x*  +  2x3  +  14x2  +  12x  +  12  x*  +  3x3  +  8x2  +  8x  +  8 

(X2  +  C2)2  -  4  &2x2  (g  -  6)3  +  (&  -  c)3  +  (C  - 

X*  +  4  6x3  +  4  {,2x2  _  c4  ■  ■  (a  -  6)  (6  -  c)  (c  -  a) 


OPERATIONS   WITH   FRACTIONS 

500  Lowest  common  denominator.  To  add  or  subtract  fractious, 
we  first  reduce  them  to  equivaleut  fractions  having  a  commoa 
denominator. 

Evidently  the  lowest  common  multiple  of  the  given  denomi- 
nators will  be  the  common  denominator  of  lowest  degree.  It 
is  therefore  called  the  lowest  common  denoviinator  (L.C.D.)  of 
the  given  fractions. 

Example.     Reduce  —,  — ,  and  —  to  a  lowest  common  denominator, 
be    ca  ab 

The  L.C.M.  of  the  given  denominators  is  abc. 

To  reduce  a /be  to  an  equivalent  fraction  having  the  denominator  abc, 
we  must  multiply  both  its  terms  by  a. 

Similarly  we  must  multiply  both  terms  of  b/ca  by  b,  and  both  terms 
of  c/ab  by  c. 

a  _  a2      6  _  ^      c  _  c2 
be      abc    ca      abc    ab      abc 

501  To  reduce  two  or  more  fractions  to  equivalent  fractions  having 
a  lowest  common  denominator,  find  the  lowest  common  miilti- 
ple  of  the  given  denominators. 

Then  in  each  fraction  replace  the  denominator  by  this  lowest 
common  multiple,  and  multiply  the  numerator  by  the  new  factor 
thus  introduced  in  the  denominator 


RATIONAL   FRACTIONS  217 

Addition  and  subtraction.     For  fractions  which  have  a  common     502 
denominator  the  rule  of  addition  and  subtraction  is  contained 
in  the  formula,  §  320,  4, 

a       b       c  _a  -\-  b  —  c 

d      d      d  d 

Hence  to  find  the  algebraic  sum  of  two  or  more  fractioit^, 

If  necessary,  reduce  them  to  a  lowest  common  de?iommato7\ 
Co7inect  the  numerators  of  the  resulting  fractions  by  the  signs 

which  connect  the  given  fractions,  and  tvrite  the  result  over  the 

common  deriominator. 

Finally,  simjdify  the  result  thus  obtained. 

This  rule  applies  when  integral  expressions  take  the  place 
of  one  or  more  of  the  fractions ;  for  such  an  expression  may 
be  regarded  as  2^.  fraction  7vhose  denominator  is  1. 

It  is  best  to  reduce  each  of  the  given  fractions  to  its  lowest 
terms,  unless  a  factor  which  would  thus  be  cancelled  occurs 
in  one  of  the  other  denominators. 

Care  should  bo  taken  that  the  expression  selected  as  the 
lowest  common  denominator  actually  is  this  denominator.  A 
frequent  mistake  is  to  treat  factors  like  a  —  b  and  b  —  a,  which 
differ  only  in  sign,  as  distinct,  and  to  introduce  both  of  them 
in  the  lowest  common  denominator. 

It  is  often  better  to  combine  the  given  fractions  by  pairs. 

Example  1.     Simplify  

a  +  b 


1  1  26 


1              26 

a-h      a'^-b'^ 

minator  is  a-  —  6^,  and  we  have 

a-b         a  +  b           26 
a2_62  '  a'-i_62      a^  -  62 

a-6  +  a  +  6-26      2a-26 

2 

a-2  -  62                  a2  ~  62 

a  +  b 

a  +  b      a-b     a^  -  b^ 


Observe  that  the  denominator  of  the  sum,  when  reduced  to  its  lowest 
terms,  may  be  but  a  factor  of  the  lowest  common  denominator. 


218  A   COLLEGE    ALGEBRA 

1  X3_3x  +  1 


Example  2.     Simplify  x 


1  —  X  X"-  —  1 


Since  the  first  denominator  is  1,  and  the  second  is  —  (x  —  1),  the  lowest 
common  denominator  is  x^  —  1.     We  therefore  have 

1         x3  -  3x  +  1  _  xs  -  X      X  +  1       x3  -  3x  +  1 

^  ~  1  -X  X2-  1  ~  X-2  -  1         X2  -  1  X2  -  1 

_  x^-x  +  x  +  l-x3  +  3x-l_    3x 

~  X2-1  ~X2-1' 

Example  3.     Simplify  -i-  +  _A_  _  _J_  _  _±- . 

Here  it  is  simpler  to  combine  the  fractions  by  pairs.    Thus, 

1 1  X  +  2  -  (X  -  2)  ^      4 

X  -  2      X  +  2  ~  X-  -  4  x2  -  4 

_^ 2     ^oX-l-(x  +  l)^         4 

x+1      x-1  x2-l  x2-l* 

_^  4      _  ^  x2  -  1  -  (x2  -  4)  ^  12 

x2-4      x-i-l"      (x2-l)(x2-4)      x*-5x2  +  4' 
x2-l  2x2  +  3x-2 


Example  4.     Simplify 


x^  +  x2  -  2  X      2  x3  +  x2  +  3  X 


By  aid  of  the  remainder  theorem,  §  415,  we  find  that  x  —  1  is  a  common 
factor  of  the  two  terms  of  the  first  fraction,  but  is  not  a  factor  of  the  second 
denominator.  We  therefore  simplify  the  first  fraction  by  cancelling  x-1 
in  both  terms,  thus  obtaining  (x  +  1)  /(x^  +  x2  +  2  x). 

By  §  465,  the  H.C.F.  of  x^  +  x^  +  2x  and  2x3  +  x2  +  3x  -  2  is  x2  +  x  +  2 ; 
and  x3  +  x2  +  2x  =  (x2  +  x  +  2)x,  2x3  +  x2  +  3x  -  2  =  (2x -  1) (x2 4  x  +  2). 

Before  reducing  to  a  common  denominator,  we  inquire  whether  2  x  —  1 
is  also  a  factor  of  the  numerator  2  x2  +  3  x  -  2.     AVe  find  that  it  is,  and 
cancelling  it,  reduce  the  second  fraction  to  (x  +  2)/(x2  -f  x  +  2). 
x2-l  2x2  +  3x-2 


Hence 


x4  +  x2-2x      2x3  +  x2  +  3x-2 

x+1         .       x+ 


X(x2  +  X+2)         X2  +  X  +  2 

x  +  l  +  x'  +  2x^  x^  +  sx^.! 

"       «3  4.  x2  +  2  X       ~  X3  +  X2  +  2  X 

503        Multiplication.     The  product  of  two  or  more  fractions  may 
be  feund  by  applying  the  following  theorem. 


RATIONAL   FRACTIONS  219 

The  product  of  two  or  more  fractions  is  the  fraction  whose 
numerator  is  the  product  of  the  numerators  of  the  given  fractions, 
and  its  denominator  the  product  of  their  denominators. 

mi-  a    c      ac 

For  the  product  of  each  member  by  hd  is  ac  (see  §  253). 

Thus,  — 6d  =  ac;  and  -• '^■6d  =  '^6-d  =  ac.  §§252,254 

hd  b    d  b      d 

In  particular,  to  multiply  a  fraction  by  an  hitegral  expres- 
sion, multiply  its  numerator  by  that  expression. 

The  fraction  ac  /hd  should  always  be  reduced  to  its  lowest 
terms  before  the  multiplications  indicated  in  its  numerator 
and  denominator  are  actually  performed. 

Example  1.     Multiply  (x-^  -  i)  /  (x3  ^  i)  by  (^  +  1)  /  (a;  -  1). 

Wehave  5^:^   x  +  1  ^  (x^  -  l)(a;  +  1)  ^  x^  +  x  +  1 

X3  +  1     X  -  1         (X3  +  1)  (X  -  1)         X2  -  X  +  1 

Example  2.     Multiply  1  -  (x  -  2)  /  (x^  +  x  -  2)  by  (x  +  2)  /x. 

_,             /            x-2\x  +  2             x2          x  +  2         X 
We  have      (  1 )  • = = 

\  x2  +  X-2/  X  X2  +  X-2  X  x-1 

Involution.     From  §  503  we  derive  the  rule  :  504 

To  raise  a  fraction  to  any  given  power,  raise  both  numerator 
and  denominator  to  that  power. 


Thus,  ■^-^       ^^ 

.6/        b" 


o 


b)  ~b"b 


For      (-1   = •••ton  factors 


to  n  factors      a» 


.6/        b    b  b-b-  •  -to  n  factors      6" 

Example.     Find  the  cube  of  —  ab"c^/efg*. 

a62c3\3  a^b'^c^ 


We  have  ( I   = 

V      efg*  )  e^Pg^"^ 

Division,     To  invert  a  fraction,  as  a  Jh,  is  to  interchange  its     505 
numerator  and  denominator.     The  fraction  h  /  a  thus  obtained 
is  called  the  reciprocal  of  a  Jh. 


220  A   COLLEGE    ALGEBRA 

To  divide  one  fraction  hy  another,  multiply  the  dividend  by 

the  reciprocal  of  the  divisor. 

a      c      a   d      ad 
Thus,  _  ^  _  =  _  .  -  =  — 

b      d      b    c      be 

For  the  product  of  each  member  by  c/d  is  a/6  (see  §  253). 

™,  a      c      c      a        J  ad   c      adc      a         ofnco  nn    mo 

Thus,  7-^3X:;  =  r;  and— •-  =  —-  =  -.        §§252,254,503 

b      d      d      b  be    d      bed      b 

In  particular,  to  divide  a  fraction  by  an  integral  expression, 
multiply  its  denominator  by  that  expression. 

Example  1.     Divide 

(X2  -  xy  +  2/2) /(x2  _  yi)  by  (X*  +  x2y2  +  yi)/{x^  -  y*). 

x2  -  xy  +  2/2      X*  +  x2?/2  +  2/*  _  x2  -  X2/  +  2/*^  X*  -  y* 


We  have 


X2  -  2/2  X*  -  2/*  X2  -  2/2  X*  +  x22/2  +  J/* 

X2  +  W2 


X2  +  XT/  +  2/2 

Example  2.    Divide  (x^  +  5  x  +  6)  /  (x  +  1)  by  x2  +  6  x  +  8. 
We  have 

»^  +  ^^  +  ^      (X2  +  6X  +  8)  -^  +  5x4-6 


x  +  1  '  (x  +  l)(x2  +  6x  +  8) 

_        (x  +  2)  (X  +  3)        _       X  +  3 
~  (X  +  1)  (X  +  2)  (X  +  4)  ~  x2  +  5x  +  4' 

506  Rational  expressions  in  general  Complex  and  continued  frac- 
tions. We  have  shown  that  the  sum,  difference,  product,  and 
quotient  of  two  fractions  are  themselves  fractions.  It  therefore 
follows  that  every  rational  expression,  §  267,  can  he  reduced  to 
the  form  of  a  simple  rational  fraction. 

No  general  rule  can  be  given  as  to  the  best  method  for  redu- 
cing a  given  rational  expression  to  its  simplest  form.  Seek  to 
avoid  all  unnecessary  steps.  In  particular  be  on  the  watch 
for  opportunities  to  reduce  fractions  to  their  lowest  terms,  and 
perform  no  multiplications  of  polynomials  until  the  reduction 
can  be  carried  no  further  without  so  doing. 

507  As  has  already  been  said,  §  495,  the  fraction  A  / B  is  called 
complex  when  A  or  B,  either  or  both,  is  fractional. 


RATIONAL   FRACTIONS  221 

In  simplifying  a  complex  fraction,  A I B^  it  is  sometimes 
better  to  divide  A  by  B  at  the  outset  by  the  rule  of  §  505, 
sometimes  better  first  to  multiply  both  A  and  B  by  the  L.C.M. 
of  all  the  denominators  in  A  and  B.  Before  taking  either  of 
these  steps  it  is  often  best  to  simplify  A  and  B  separately. 

a  +  &  ,  ,\    I  /  a  —  h 


E.amp.el.    Simplify  (»^;+l)/("^°  +  l). 


We  have 


a  +  h  a  +  h  +  a  —  h        2a 

a  —  b  a  —  b  a  —  b 


a-b  a -b  +  a  +  b        2a 

a+6  a+b  a+6 

_    2a     a  +  6  _  a  +  b 

~  a  —  b      2a        a  —  b 

Observe  that  when  the  terms  of  a  complex  fraction  are  simple  fractions 
we  may  cancel  any  factors  which  are  common  to  the  numerators  or  to 
the  denominators  of  these  fractions.  Thus,  in  the  third  expression  above, 
we  may  cancel  2  a. 

Example  2.     Simplify  (^^  -  ^^)  /  (-^^  +  -A_  )  . 

We  may  proceed  as  in  Ex.  1 ;  but  a  simpler  method  is  to  begin  by 
multiplying  both  terms  by  (a  +  b){a-b).     We  thus  obtain 

a            b 
a  -  b~  a  +  b     a{a  +  b)  -  b{a  -  b)  _  a"^  +  ab  -  ba  +  b"^  _  ^ 
"T~      "~&         a{a  -  b)  +  b{a  +  b)  ~  a^  -  ab  +  ba  +  b^ 
a  +  b      a  —  b 
Example  3.     Simplify  ■ 

"■^^ 

Working  from  the  bottom  upwards,  we  have 

a  a  a(d/  +  e)  adf  +  ae 


c         ,    ,      cf         b(df+e)  +  cf      bdf+be  +  cf 

Complex  fractions  like  that  in  Ex.  3  are  called  continued 
fractions. 


222  A   COLLEGE   ALGEBRA 

EXERCISE  XXV 

Simplify  the  following  expressions. 


2a-Sb      2a  +  3b      ia^-9b^ 

X  +   1         X2  -  1         X3  +  1 
1  1 


x2-3x  +  2      x2-5x  +  6      x2-4x  +  3 
x+1  x+2_  x+3 


(X  -  1)  (X  -  2)       (2  -  X)  (X  -  3)       (3  -  X)  (1  -  x) 

5.   -i ^  +  -i '-. 

X  +  b      X  +  c      X  —  b      X  —  c 


(a  -b){a-  c)      {b  -c){b-  a)      (c  -a){c-  b) 

^        yz  (x  +  g)       ^       zx  {y  +  a)       ^       xy  (z  +  a) 
{x-y){x-z)      {y-z){y-x)      (z  -  x)  (z  -  j/) 

1  8x*-33x  2x  +  6 


x  + 


3-2x        8x3- 27        4x2  +  6x  + 

1\2 


(.,i)%(,,i)-,(.,,i)-_(.,^X,,>)(.,,i) 


(a  +  6)^-c3      {6  +  c)3-a3      (c  +  a)^  -  b' 

a  +  6  —  c  5  +  c  —  a  c  +  a  —  6 

x2-4  3x2-14x-5 


11. 


x3-3x2-x  +  6      3x3-2x2-10x 


12.    . ' + ' +  ' 

x4_4x2_a;  +  2      2x*-3x3-5x2  +  7x-2      2x*  +  3x3-2x2-2x  +  l 

-(--i)-(''4)-  -a-^D<'"->- 

,^    «2_  6x +  6   x2  + 7x  +  12       X2  +  X-6 
15. 


x2  +  3x-4   x2-8x  +  15      x2-4x-5 
ifi    1      Ji       rx-1      1/x-l      (x-2)(x-3)\-t1 


RATIONAL   FRACTIONS  223 


^^_    ax  +  x^_26x^3_  ^^_   (.._^._,.  +  2,z)^^-^  +  ^ 


19. 


2  6  -  ex      (a  +  x)2  "  '     X  -  y  -  z 

/a  +  b      a^  +  b^\/a  +  b  a"^  +  b^\ 

1 1_      1 1_  a^b        J__j_ 

20    ^      y  +  zy      x  +  z  ^^     b      a        a*      6* 


1  1  1  1  a      b      /I    .  1\2 

z      y  +  z      y      x  +  z  b 


(„-.) 


22.    ^^=^ 23.    x  +  ^ 


x-2- 


x  +  1 


x-1  «  +  !_ 

x-2"  »' 


INDETERMINATE  FORMS 


Limits.  Suppose  the  variable  x  to  be  taking  successively  508 
the  values  1/2,  3/4,  7/8,  15/16,  and  so  on  without  end; 
then  evidently  x  is  approaching  the  value  1,  and  in  such  a 
manner  that  the  difference  1  —  x  will  ultimately  become  and 
remain  less  than  every  positive  number  that  we  can  assign,  no 
matter  how  small  that  number  may  be.  We  indicate  all  this 
by  saying  that  as  x  runs  through  the  never-ending  sequence  of 
values  1/2,  3/4,  7/8,  15/16,  •••,  it  apjwoaches  1  as  limit. 

And  in  general,  if  x  denote  a  variable  ivhich  is  supposed  to  be 
rimning  through  some  given  but  never-ending  sequence  of  values, 
and  if  there  be  a  7iumber  a  s^ich  that  the  difference  a  —  x  will 
ultimately  become  and  remain  numerically  less  than  every  posi- 
tive number  that  we  can  assign,  we  say  that  x  approaches  this 
number  a  as  limit. 

To  indicate  that  x  is  approaching  the  limit  a,  we  write 
X  =  a,  or  a  =  lim  x. 

It  will  be  noticed  that  the  word  variable  has  here  a  more  restricted 
meaning  than  in  §  242. 

Whether  or  not  a  variable  of  the  kind  here  under  consideration 
approaches  a  limit  depends  on  the  sequence  of  values  through  which 


224  A   COLLEGE   ALGEBRA 

it  is  supposed  to  be  running.     Thus,  if  the  sequence  be  1,  2,  1,  2,  ■  •  • ,  the 
variable  will  not  approach  a  limit. 

A  full  discassion  of  variables  and  limits  will  be  found  in  §§  187-205, 
which  the  student  is  advised  to  read,  at  least  in  part,  in  this  connection. 

509  Theorems  respecting  limits.  In  §  203  it  is  proved  that  if  the 
variables  x  and  y  approach  limits,  then  their  sum,  difference, 
product,  and  quotient  also  approach  limits,  and 

lim  (x  -\-  y)  =  lim  x  +  lim  y. 
lim  (x  —  y)  =  lim  x  —  lim  y. 
lim  xy         =  lim  x  ■  lim  y. 

lim  -  =  t: ,  unless  lim  ?/  =  0. 

y  hm  y 

From  these  theorems  it  follows  that  if  F(x)  denote  any 
given  rational  function  of  x,  and  F(a)  its  value  when  x  =  a, 
then  F(x)  will  approach  F(a)  as  limit  ivhenever  x  approaches 
a  as  limit,  that  is, 

limF(a-)  =  F{a), 

where  ^^^^  F(.r)  is  read  "limit  of  F(x'),  as  x  approaches  a." 
Thus,  ^™  (2x2  -  3x  +  1)  =  2 a2  _  3 a  +  L 

510  Infinity.  Evidently  if  x  be  made  to  run  through  the  never 
ending  sequence  1,  2,  3,  4,  •  •  • ,  it  will  ultimately  become  and 
remain  greater  than  every  number  that  we  can  assign. 

A  variable  x  %vhich  will  thiis  ultimately  become  and  remain 
numerically  greater  than  every  number  that  we  can  assign  is 
said  to  approach  infinity. 

For  the  word  infi7iity  we  employ  the  symbol  oo,  and, 
to  indicate  that  x  is  approaching  infinity,  write  x  =  <X),  or 
lim  X  =  00. 

511  Note.  It  is  important  to  notice  that  oo,  as  thus  defined,  does  not  denote 
a  definite  number,  and  that  the  rules  for  reckoning  with  numbers  do  not 
apply  to  it.     Illustrations  of  this  will  be  found  below. 


RATIONAL    FRACTIONS  225 

The  phrase  "x  is  approaching  infinity  "  is  merely  an  abbreviation  for 
"X  is  a  variable  which  will  ultimately  become  and  remain  greater  than 
every  number  that  we  can  assign." 

When,  as  is  sometimes  convenient,  we  write  lim  a;  =  oo,  of  course  the 
word  limit  does  not  have  the  definite  meaning  given  it  in  §  508. 

Theorem.      Given  any  fraction  whose  numerator  is  a  constant     512 
and  its  denominator  a  variable. 

If  the  denominator  approach  0  as  limit,  the  fraction  will 
approach  oo;  and  if  the  denominator  approach  go,  the  fraction 
will  approach  0  as  limit. 

Thus,  consider  the  fraction  1  /x. 

If  X  approach  0  by  running  through  the  sequence  of  values  1,  .1,  .01, 
.001,  •  •  •,  then  1/x  will  run  through  the  sequence  1,  10,  100,  1000,  •  •  •, 
and  will  therefore  approach  oo. 

And  if  X  approach  oo  by  ruiuiing  through  the  sequence  1,  10,  100, 
1000,  •••,  then  1/x  will  run  through  the  sequence  1,  ,1,  .01,  .OOl,---, 
and  will  therefore  approach  0  as  limit. 

And  so  in  general. 

Indeterminate  forms.  A  rational  fraction  of  the  form  513 
/(.r)/<^(.r)  has  a  definite  value  for  any  given  value  of  x 
except  one  for  which  <^  (x)  =  0.  But  when  <^  (.r)  =  0,  the  frac- 
tion takes  one  of  the  forms  0/0  or  o/O,  which  are  arithmet- 
ically meaningless,  §  103.  It  is  convenient,  nevertheless,  to 
assign  a  meaning  to  the  fraction  in  both  of  these  cases. 

The  form  0/0.      The  fraction  {x"^  -  1)  /  {x  -  1)   takes  the     514 
form  0/0  when  x  =  \. 

Now,  except  when  a;  =  1,  we  can  divide  a*^  —  1  by  a:  —  1, 

and  have 

(a-2-l)/(a;-l)  =  a^  +  1. 

This  is  true  however  little  x  may  differ  from  1.     Hence  if, 

without  actually  giving  x  the  value  1,  we  make  it  approach  1 

as  limit,  we  shall  have 

lim  (a;2  _  1)  /(a;  -  1)  =  lim  (a-  +  1)  =  2. 

Thus,  while  the  rules  of  reckoning  give  us  no  meaning  for 
(x^  —  1)  /  (x  —  1)  when  a;  =  1,  they  enable  us  to  prove  that 


226  A    COLLEGE    ALGEBRA 

this  fraction  always  approaches  the  definite  limit  2  when  x 
approaches  1  as  limit. 

Now  we  have  just  shown,  §  509,  that  F {a) —  ^}^^F {x) 
whenever  F{x)  represents  a  rational  function  and  F(a)  has 
a  meaning.  It  is  therefore  convenient  when,  as  here,  the 
rules  of  reckoning  give  us  no  meaning  for  F{(x),  to  assign  to 
it  the  value  ^™F(a;);  in  other  words,  to  make  the  formula 
F{a)  =  li^i^  F{x)  our  definition  of  F(a). 

We  therefore  give  (x-  —  1)  /  (x  —  1)  the  value  2  when  x  —  \. 
We  then  have  {x?  —  V)  J  {x  —  V)  =  x  ■\-  1  for  all  values  of  a-, 
the  value  1  included. 

And  for  a  like  reason,  to  every  fraction  which  can  be  written 
in  the  form  {x  —  a)f(x)/(x  —  a)<f)(x),  where  f(x)  and  cl)(x) 
are  integral  and  cf>  (x)  is  not  divisible  by  x  —  a,  we  assign  the 
value /(a) /^  (a)  when  x  =  a,  and  so  have 

(x  -  a) fix)  ^  f(x) 
(x  —  a)  (f>  (x)       <f>  (x) 

for  all  values  of  x,  the  value  a  included. 
515         The  form  a/0.     The  fraction  1/x  takes  the  form  1/0  when 
x  =  0. 

While  we  cannot  divide  1  by  0,  we  can  divide  1  by  a  value 
of  X  which  differs  as  little  as  we  please  from  0.  Moreover  we 
have  shown,  §  512,  that  if  x  be  made  to  approach  0  as  limit, 
then  1/x  will  approach  oo. 

We  therefore  assign  to  1/0,  and  in  general  to  a/0,  when 
a=^  0,  the  "  value  "  oo,  writing 

a 

^  =  00. 

And  for  a  like  reason,  to  every  fraction  of  the  form 

/(x)/(x-«)<^(,r), 

where /(cr)  and  <f>(x)  are  integral  and /(.t)  is  not  divisible  by 
X  —  a,  we  assign  the  "value"  oo  when  x  =  a;   our  meaning 


RATIONAL   FRACTIONS  227 

being  that  the  fraction  will  always  approach  oo  when  x  is 
made  to  approach  a  as  limit. 

Conclusion  as  to  the  values  of  a  fraction.     From  §§  514,  515     516 
we  draw  the  following  conclusions  regarding  a  simple  fraction 
of  the  form  f{x)l^ix). 

1.  If /(x)/<^(a:)  is  in  its  lowest  terms,  it  will  vanish  for 
values  of  x  which  make  its  numerator  f  (x)  vanish,  and  become 
infinite  for  values  of  x  which  make  its  denoniinator  <^  (x)  vanish. 
For  all  other  finite  values  of  x  it  has  a  value  different  from 
both  0  and  co. 

2.  But  if  f(x)  and  <^  (x)  have  the  common  factor  x  —  a, 
and  /(ic)  contains  this  factor  m  times  and  <^(x)  contains  it 
n  times,  then  f(x)/cf)(x)  will  vanish  for  x  =  a  when  m  >  n, 
become  infinite  when  in  <  u,  and  have  a  value  different  from 
both  0  and  co  when  m  —  n. 

Thus  when  x  =  2,  we  have 

a;-2^^   ^±2  =  00    ^^^z31  =  o      ^^-^^  (X  -  2)2  ^1 

x  +  1        'x-2      '^'x(x-2)        'x(x-2)2      '^'  x(x-2)2      2' 

The  form  00/00.     It  is  often  important  to  know  what  limit     517 
the  value  of  a  fraction /(aj)/^  (a;)  approaches  when  x  is  indefi- 
nitely increased,  that  is,  when  x  =  <x>. 

Consider  the  following  examples. 

We  have  shown,  §  512,  that  1/x,  1  /x^,    •  ■  =  0,  when  x  =  00. 

XT  u  .  X2-X  +  3        1  -  1/X  + 3/x2   .    ^  ^^  ,,^ 

Hence,  when  x  =  oo,  —  = ■ — - —  =  1/2,  (1) 

2x2  +  x-4      2  +  1/X-4/X2  ^' 


X  +  2  1  +  2/x 


x2  +  x  +  5      X  +  1  +  5/X 
x2  +  x-7      X  +  1-7/X 


=  0,  (2) 

=  00.  (3) 


2x  +  3  2  +  3/x 

And  in  general,  when  x  =  <x>,  the  fraction 

{a^x^  +  aia:"'-^  +  •  •  •  +  a,„) /  {b^x-  +  Wx^-^  +  . . .  +  ^, J 
approaches  the  limit  ao/bo,  if,  as  in  (1),  the  degrees  of  numer- 
ator and  denominator  are  the  same ;  the  limit  0,  if,  as  in  (2), 


228  A   COLLEGE   ALGEBRA 

the  degree  of  the  denominator  is  the  greater ;  the  limit  oo,  if, 
as  in  (3),  the  degree  of  the  numerator  is  the  greater. 

And  in  each  case  the  limit  is  called  the  "value  which  the 
fraction  takes  when  x  =  oo,"  that  is,  when  the  fraction  itself 
assumes  the  indeterminate  form  oo/co. 
518  The  forms  0  •  oo  and  oo  —  oo.  A  rational  function  of  x  may 
take  one  of  the  indeterminate  forms  0  •  oo  or  oo  —  oc  for  some 
particular  value  of  x.  But  the  expression  can  then  be  reduced 
to  one  which  will  take  one  of  the  forms  0/0,  a/0,  or  oo/oo 
already  considered. 

1.  Thus,  (x2  —  1)  • takes  the  form  0  •  co  when  x=  I.     But,  except 

"^ "  ^  1         x2  -  1 

when  X  =  1,  we  have  (x^  —  I) = ,  and  therefore 

X  —  1       X  —  1 

lim  r/a;2  _  1)  .  _J_'\  =  lim  ^^^ll  =  lim  (x  +  1)  =  2. 

Hence  we  assign  to  the  given  expression  the  value  2,  when  x  =  1. 

1  2 

2.  Again, takes   tlie  form   oo  —  oo  when  x  =  0.      But, 

X      x(x  +  2)     ^  2 

except  when  x  =  0,  we  have = ,  and  therefore 

X      X  (X  +  2)      X  (x  +  2) 


lim  n ?_"1  =  lim 

'^^Olx         X(X  +  2)J  =^=0; 


lim      1       _1 


x(x  +  2)J       =^-0x{x  +  2)      ^-"x  +  2      2 
Hence  we  assign  to  the  given  expression  the  value  1/2  when  x  =  0. 

519  General  conclusion.  Therefore,  if  a  given  function  of  a  single 
variable,  as  F(x),  assumes  an  indeterminate  form  when  x  =  a, 
proceed  as  follows : 

Reduce  the  exj^ression  to  its  simj)Iest  form,  and  then  find 
what  limit  its  value  approaches  when  x  is  made  to  approach 
a  as  limit.  Call  this  limit  the  value  which  the  function  has 
when  X  =  a. 

520  Note.  This  method  is  restricted  to  functions  of  a  single  variable,  as 
F{x).  For  the  reason  that  the  method  yields  definite  results  is  this  :  the 
value  of  ''™  F  (x)  depends  solely  on  the  value  of  a  and  not  on  the  values 
which  X  may  take  in  approaching  a ;  and  the  like  is  not  true  of  functions 
of  more  than  one  variable. 


RATIONAL    FRACTIONS  229 

Thus,  suppose  that  x  and  y  are  unrelated  variables,  and  consider  the 
fraction  x/y  when  x  =  0  and  y  =  0. 

The  limit,  if  any,  which  x/y  will  approach  when  x  =  0  and  y  =  0, 
depends  on  the  sequences  of  values  through  which  x  and  y  may  run. 

For  example,  a  variable  will  approach  0  as  limit  if  it  runs  through 
either  of  the  following  sequences  : 

1/2,  1/3,  1/4,  ■••     (1);         1/2^  1/3-:,  1/42,  •••     (2) 

If  X  runs  through  (1),  and  y  through  (2),  then  x/y  will  run  through 
the  sequence  2,  3,  4,  •  ■  • ,  and  approach  oo.  But  if  x  runs  through  (2), 
and  y  through  (1),  then  x/y  will  run  through  the  sequence  1/2,  1/3, 
1  /4,  •  •  • ,  and  approach  0. 

Therefore,  if  x  and  y  are  unrelated  variables,  we  regard  x/y  as  abso- 
lutely indeterminate  when  x  =  0  and  y  =  0.     And  so  in  general. 

Infinity  in  relation  to  the  rules  of  reckoning.      If  we  take     521 
infinite  values  of  the  letters  into  account,  we  must  state  the 
rules  of  §§  249,  251,  253  as  follows : 

1.  a  •  0  =  0,  unless  a.  =  oo. 

2.  If         ac  =  he,         then  a  =  b,  unless  c  =  0  or  oo. 

3.  li  a  -{-  c  =  b  +  c,  then  a  =  b,  unless  c  =  oo. 

It  is  important  to  keep  these  exceptional  cases  in  mind 
when  applying  the  rules  to  the  solution  of  equations. 

Thus,  consider  the  product x  —  1.     When  x  =  1,  the  second 

x^  —  1 
factor,  X  —  1,  is  0  ;  but  as  the  first  factor,  1  /(x^  —  1),  is  then  oo,  it  does 
not  follow  that  the  product  is  0.     The  product  is  1/2  in  fact,  §  518. 

Infinite  roots  of  equations.      Instead  of  saying,  as  we  have     522 
been  doing,  that  the  equation  x  -\-  2  =  x  -\-  S  and  other  simple 
equations  which  will  reduce  to  the  form  0-x  =  b,  have  no 
root,  we  sometimes  say  that  the^/  have  the  root  oo. 

For  however  small  a  may  be,  if  not  actually  0,  ax  =  b 
has  the  root  b/a.  And  if,  keeping  b  constant  and  different 
from  0,  we  make  a  approach  0  as  limit,  b/a  will  approach  oo, 
§  512.  In  other  words,  a,s  ax  =  b  approaches  the  form  Ox  —  b, 
its  root  b/a  approaches  the  value  oo.     It  is  therefore  quite  in 


230  A   COLLEGE   ALGEBRA 

agreement  with  the  practice  explained  in  §  515  to  say  that 
when  ax  =  b  has  the  form  Ox  =  h,  it  has  the  root  oo. 

Observe  that  if  we  regard  a;  +  2  —  x  +  3asa  true  equation  whose  root 
is  00,  we  are  not  driven  to  the  absurd  condusion  that  2  =  3.  For  since 
a;  =  CO  we  have  no  right  to  infer  that  the  result  of  subtracting  x  from  both 
members  is  a  true  equation,  §  521,  3. 

523  Infinite  solutions  of  simultaneous  equations.  In  like  manner, 
instead  of  saying  of  a  system  of  inconsistent  simple  equations, 
§  377,  2,  §  394,  2,  that  it  has  no  solution,  we  sometimes  say 
that  it  has  an  infinite  solution;  for  from  such  a  s^^stem  we 
can  obtain  by  elimination  a  single  equation  of  the  form  Ox  =  b, 
and,  by  §  522,  this  equation  has  the  root  oo. 

Thus,  we  may  say  that  the  pair  of  inconsistent  equations  y  —  x  =  0  (1), 
y  —  a;  =  1  (2)  has  an  infinite  solution. 

Observe  that  this  pair  (1),  (2)  is  the  limiting  case,  as  m  =  1,  of  the  pair 

y  -  mx  =  0  (3),  y-x  =  l  (2). 
The  solution  of  the  pair  (3),  (2)  is 

X  =  l/(m  —  1),  y  =  m/ {in  —  1), 
and  when  m  =  1,  both  l/(m  —  1)  and  mf  {m  —  \)  approach  infinity. 

The  same  thing  may  be  shown  by  the  graphical  method,  §§  386,  387. 
For,  when  m  =  1,  the  graph  of  (3)  approaches  parallelism  with  that  of 
(1),  and  the  point  of  intersection  of  the  two  graphs  recedes  to  an  infinite 
distance  in  the  plane. 

EXERCISE  XXVI 

Assign  the  appropriate  values  to  the  following  expressions. 

2.  2.    1  when  x  =  1. 

x3  -  2  X  +  1 

x2-2ax  +  a2 

1.  4. 1  when  x  =  a. 

x2-(a+6)x  +  a6 

when  X  =  —  2. 

when  X  =  1. 
x3-3x2  +  3x-l 


X2- 

5x  +  6 

X2- 

6x  +  8' 

X 

2-1 

X2- 

2x  +  l 

(3 

X  +  1)  (X  +  2)2 

(X2- 

-  4)  (x2  + 

3x  +  2) 

X3 

-X2-X 

+  1       . 

RATIONAL   FRACTIONS  231 


^      Sz^-x  +  5  ^  x^  +  1^  _3x_^   (2.^  +  l)(x3-5)^  when  x  =  oo. 
*    2  x2  +  6  X  -  7         X        x2  +  1       (x*  +  1)  (X  -  6) 

-X  —  1  X  —  2  ,  „ 

8. 1  when  x  =  3. 

x2  -  9      X  (X  -  3) 

9. 1 5  when  x  =  1. 

X  —  1      X  (x  —  1) 

X2  +  ^  ^-^ 

10.    ^ ,  whenx  =  2.  11.   .  when  x  =  oo. 

2       X-  1  3x  +  l 

"^    ^  X  -  2  x2  +  1 


FRACTIONAL   EQUATIONS 

On  solving  a  fractional  equation.     Any  given  fractional  eqiia-    524 
tion  may  be  transformed  into  one  which  is  integral  by  multi- 
plying both  its  members  by  D,  the  lowest  common  denominator 
of  all  its  fractions.     We  call  this  process  clearing  the  equation 
of  fractions. 

It  follows  from  §§  341,  342  that  the  integral  equation 
which  is  thus  derived  will  have  all  the  roots  of  the  given 
equation,  and,  if  it  has  any  roots  besides  these,  that  they 
must  be  roots  of  the  equation  D  =  0  and  so  may  readily  be 
detected  and  rejected. 

Example  1.     Solve  ?  +  — ^^^  =  0.  (1) 

xx-lx(x-l) 

Clear  of  fractions  by  multiplying  by  D  =  x  (x  —  1). 

We  obtain  3(x  -  1)  +  6x  -  (x  +  13)  =  0.  (2) 

Solving  (2),  X  =  2.  (3) 

Therefore,  since  2  is  not  a  root  of  D  =  x  (x  -  1)  =  0,  it  is  a  root  of  (1), 

and  the  only  root. 


Example  2.     Solve  ?  +      \         ?^^,,  =  0- 
X      x-1      x{x-l) 

(1) 

Clearing  of  fractions,  3  (x  -  1)  +  6  x  -  (x  +  5)  =  0. 

(2) 

Solving  (2),                                                      a;  =  1- 

(3) 

232  A   COLLEGE    ALGEBRA 

As  1  is  a  root  of  D  =  a;(x  -  1)  =  0,  it  is  not  a  root  of  (1).  In  fact, 
when  X  =  1,  the  first  member  of  (1)  has  the  form  3  +  6/0-G/O;  and  by 
§  518  we  find  that  its  value  is  8,  not  0. 

Hence  (1)  has  no  root. 

We  may  sum  vip  the  method  thus  illustrated  in  the  rule : 

525  To  solve  a  fractional  equation,  clear  of  fractions  hy  multiply- 
ing both  members  by  D,  the  lowest  common  denominator  of  all 
the  fractions. 

Solve  the  resulting  integral  eqxiation. 

The  roots  of  this  equation  —  except  those,  if  any,  for  which 
D  vanishes  —  are  the  roots  of  the  given  equation. 

526  Note.     We  may  also  establish  this  rule  as  follows  : 

Let  N /D  =  0  represent  the  result  of  collecting  all  the  terms  of  the 
given  equation  in  one  member  and  adding  them  ;  then  N  =  0  will  be  the 
integral  equation  obtained  by  clearing  of  fractions. 

1.  If  iV/D  is  in  its  lowest  terms,  the  roots  of  iV/D  =  0  and  i\r  =  0  are 
the  same  ;  for  a  fraction  in  its  lowest  terms  vanishes  when  its  numerator 
vanishes,  and  then  only,  §  516. 

3  6  x+,  13        8{x-2)      N 

Thus,  in  §  524,  Ex.  1,  -  + ^  =  ~ .{  =  7:- 

'       ^  'x      x-1      x(x-l)      x(x-l)      D 

Here  N/D  is  in  its  lowest  terms  and  the  root  of  iV/D  =  0  is  the  same 
as  the  root  of  N  —  0,  namely,  2. 

2.  If  N/J)  is  not  in  its  loioest  terms,  iV  =  0  will  have  roots  which 
N/D  =  0  does  not  have,  namely,  the  roots  for  which  the  factor  common 
to  N  and  D  vanishes. 

3  6  x  +  5         8(x-l)iV 

Thus,  in  §  524,  Ex.  2,  -  + =  -^ ^  =  -• 

'       *  'x      x-1      x(x-l)      x(x-l)      D 

Here  N/D  is  not  in  its  lowest  terms,  and  the  root  of  iV  =  0,  namely  1, 
is  not  a  root  of  N/D  =  0;  for  when  x  =  1,  iV/D  =  8,  §  514. 

Evidently  1.  is  the  general  case  and  2.  is  exceptional. 

3  6  X  +  a        ^ 

Thus,  consider  the  equation      -\ ; —  =  v. 

X      x-1      x(x-l) 

Here  N/D  =  [8x  -  (a  +  3)]/x(x  -  1),  and  this  is  in  its  lowest  terms 
except  when  u  =  5  or  —  3. 


RATIONAL   FRACTIONS  233 

3.    It  must  not  be  inferred  from  what  has  just  been  said  that  the  given 

equation  will  never  be  satisfied  by  a  root  of  iV"  =  0  which  is  also  a  root  of 

X»  =  0. 

X  2  1 

Thus,  consider  the  equation =  0. 

X  —  1      X      X  (x  —  1) 

Here  N/D  =  (x  —  l)2/x(x  —  1)  and,  by  §  516,  this  expression  vanishes 
when  X  =  1.  But  observe  that  N  =  (x  —  1)^  =  0  has  the  root  1  a  greater 
number  of  times  than  D  =  x{x  —  1)  —  0  has  this  root. 

In  applying  the  rule  of  §  525,  care  must  be  taken  not  to     527 
introduce  extraneous  factors  in  the  expression  selected  as  the 
lowest  common  denominator. 

If  any  fraction  in  the  equation  is  not  in  its  lowest  terms, 
begin  by  simplifying  this  fraction,  unless  the  factors  thus 
cancelled  occur  in  other  denominators. 

Before  clearing  of  fractions  it  is  sometimes  best  to  combine 
certain  of  the  fractions,  or  to  reduce  certain  of  them  to  mixed 
expressions. 

6  X  +  5  x2  11 


Example  1.     Solve 


!x  +  15      6x-2x2       5 


Here  the  terms  of  the  first  fraction  have  the  common  factor  x  —  5, 
and  those  of  the  second  the  common  factor  x.  Cancelling  these  factors, 
we  have 

x-1  x  11         x-1,         X  11 


x-3      6-2x" 

5 

x-3 

2(x-3)" 

5 

Clearing 

of  fractions. 

lOx- 

-10  +  5x: 

=  22x- 

Solving, 

X  -- 

=  8. 

Example 

2.     Solve  ^  +  V 
x  +  2 

^x  +  6 
x  +  7 

x  +  2 
x  +  3 

x  +  5_ 
x  +  6 

Reducing 

;  each  fraction  to 

a  mixed 

expression  and  simplifying. 

.^- 

1 
x  +  7 

^- 

1 
x  +  6 

Transposing  so  that  the  terms  in  each  member  may  be  connected  by 
minus  signs, 

_J 1__^ 1_ 

x  +  2      x  +  3~x  +  6      x  +  7* 


234  A   COLLEGE    ALGEBRA 

Combining  the  terms  of  each  member  separately, 
1  1 


X-  +  bx  +  6      x^  -f  13 X  +  42 
■     Clearing  of  fractions,  x^  +  13  x  +  42  =  x^  +  5  x  +  6. 

Solving,  x  =  -9/2. 

The  given  equation  may  be  solved  by  clearing  it  of  fractions  as  it 
stands,  but  that  method  is  much  more  laborious. 

528  Simultaneous  fractional  equations.  The  general  method  of 
solving  such  a  system  is  to  clear  the  several  equations  of  frac- 
tions and  then  to  find,  if  possible,  the  solutions  of  the  result- 
ing system  of  integral  equations.  The  solutions  thus  found 
—  except  those,  if  any,  which  make  denominators  in  the 
given  equations  vanish  —  are  the  solutions  of  these  equations, 
§371. 

But  if  the  equations  are  of  the  form  described  in  §  379,  or 
if  they  can  be  reduced  to  this  form,  they  should  be  solved  by 
the  method  explained  in  that  section. 

Example  1.     Solve  the  following  pair  of  equations  for  x,  y. 

y  —  2      2/  —  4  xy  —  2x      iy  —  2y'^      xy 

Clearing  both  equations  of  fractions  and  simplifying,  v^'e  obtain 

x-?/  +  l  =  0,  X  +  2y  -  8  =  0. 
Solving  X  =  2,  2/  =  3. 

Therefore,  since  none  of  the  denominators  in  the  given  equations 
vanish  v?hen  x  =  2  and  y  =  3,  these  equations  have  the  solution  x  =  2, 
2/ =  3. 

Example  2.     Solve  the  follow^ing  system  for  x,  y,  z. 

^-±1=^^,  -y^  =  -l  2iz  +  x)  +  xz  =  0. 
xy        G    y  +  z         2       ^         ' 

These  equations  can  be  reduced  to  the  form,  §  379, 

1       1_5     1      1__2     1      1__1 

x      y      6    y      z  3    z      x  2 

Solving  for  1 /x,  l/?y,  1/z,  we  find  1/x  =  1/2,  1/?/ =  1/3,  1/z  =  -  1. 
Hence  the  required  solution  is  x  =  2,  y  —  3,  z  =  —  1. 


RATIONAL   FRACTIONS  236 


EXERCISE   XXVn 

Solve  the  following  equations  for  x. 

1. 

6X-1      4x-7^^ 
3x  +  2      2x-5 

? 

6x                8            1 

5x-l      3-15X      6 

s 

•4              1                    4 

x-2      x-4      x2-6x  +  8 

+  ' 


2x  +  3      x-5      2x2-7x-15 
1  2 


(x  +  l)(x-3)       (x-3)(x  +  2)       (x  +  2)(x  +  l) 

2 2  3 

x2-l      x2  +  4x-5      x2  +  6x  +  5~ 
x+  1  2x  5 


=  0. 


3x+l      5-6x      5  +  9x-18x2 

g    _^+  g     ^      x  +  6     ^a  +  6        g    x3  +  1      a;^  -  1  _ 
"    6  (X  +  &)      a  (X  +  a)         a6    '         '    x  +  1       x  -  1  ~ 

10    ^''  +  ^^  +  1  +       =^  -  1 


x2  +  5x  +  4      x2  +  3x-4 
^^    x-8      x-9_x  +  7      x  +  2 
'x-3      x-4~x  +  8      x  +  3* 

12    ^  +  '^  .  ^  +  9  _  a:  +  10      x  +  6 
"x  +  e      x  +  S^x  +  O       x  +  5* 

„     x3  +  2      x3-2  15 

16. =  4  X. 

x-2       x  +  2       x2-4 

14        1  ^-^    I  ^^'  +  ^^^0 

■     X  -  1         X2  -  1  1  -  X* 

15.    -J-  +  -1^±A L_^o 

x3-8      2x2  +  4x  +  8      x-2 

ax  +  c      bx  +  d  . 

16. 1 =  a  +  b. 

X  —  p       X  —  q 


236 


A   COLLEGE    ALGEBRA 


17. 


X-  1 


X  —  8      X-  +  X 


X  +  2 


2  x2  -  X  +  7 


„     x2-ax  +  26x-2a6      62-x2         3  c2 

18.      ; 1 — —  + 


X  —  26      X  —  2c 
(x  -  6)2  (X  -  c)2 


19     (^  -  '^y-    I 

(X  -  6)  (X  -  c)      (X  -  c)  (X  -  a)   ■   (X  -  a)  (x  -  6) 

20   ^^^±^-^iiA_^il2^^ 

X2  +  X         X2  -  1         X2  -  X 

21.    -^  +  -^-^+1  =  0. 
x  +  2      x-2      x2-4 

Solve  the  following  for  x  and  y,  or  for  x,  ?/,  and  z. 


22. 


3x  +  y  -  1  _  6 

X-2/  +  2   ~  7' 
X  +  9  _x  +  3 
[2/  +  4~^3" 


24. 


xy 
X  +  y 


'  +  z 


a, 


23. 


25.    <^ 


PARTIAL  FRACTIONS 


529  It  follows  from  §  506  that  every  rational  function  of  a 
single  variable,  as  x,  can  be  reduced  to  the  form  of  an  inte- 
gral function,  or  a  proper  fraction,  or  the  sum  of  an  integral 
function  and  a  proper  fraction. 

For  certain  purposes  it  is  useful  to  carry  this  reduction 
further  and,  whe7i  a  proper  fraction  A/B  is  given,  find  the 
simplest  set  of  fractiovs  of  which  A/B  is  the  sum.  The 
method  depends  on  the  following  theorems  in  wMch  the 
letters  A,  B,  P,  Q,  and  so  on,  denote  rational  integral  func- 
tions of  X. 


RATIONAL   FRACTIONS  237 

Theorem  1.      The  sum  and  the  difference  of  two  proper  frac-     530 
tions    A/B  and  C/D  are  themselves  proper  fractions. 

A   ,   C      AD±BC 

For  —  ±  —  = 

B      D  ED 

Since  A  is  of  lower  degree  than  B,  AD  is  of  lower  degree 
than  BD. 

And  since  C  is  of  lower  degree  than  D,  BC  is  of  lower 
degree  than  BD. 

Hence  AD  ±  BC  is  of  lower  degree  than  BD. 

Theorem  2.     Let  1  and  V  denote  integral  f mictions,  and  A./ 'B     531 
and  A.' /W  projier  fractions. 

If  I  +  A/B  =  I'  +  A'/B',  then  1  =  1'  and  A/B  =  A'/B'. 
For,  by  hypothesis,    I  —  f  =  A'/B'  —  A/B. 

But  I  —  I'  denotes  an  integral  function  (or  0),  and,  §  530, 
A'/B'  —  A/B  denotes  a  proper  fraction  (or  0). 

Therefore,  since  an  integral  function  cannot  be  identically 
equal  to  a  proper  fraction,  we  have 

/  -  /'  =  0  and  A'/B'  -  A/B=  0, 
or  I  =r  and  A/B  =  A  '/B'. 

Theorem  3.      Let  A/ FQ  denote  a  proper  fraction  who.ie  denom-     532 
inator  has  been  separated  into  two  factors,  P  a7id  Q,  ivhich  are 
jjrime  to  one  another. 

This  fraction  can  be  reduced  to  a  sum  of  two  proper  fractions 
of  the  forms,  B/P  and  C/Q. 

For,  since  Q  is  prime  to  P,  we  can  find,  §  479,  two  integral 

functions  M  and  N,  such  that 

1  =  il/Q  +  NP,  and  therefore  A  =  AMQ  +  ANP. 

A        AMQ + ANP      AM      AN 

Hence  —  = = (i) 

PQ  PQ  P  Q  ^  ^ 

If  AM/P  and  AN / Q  are  proper  fractions,  our  theorem  is 
already  demonstrated. 


238  A   COLLEGE   ALGEBRA 

If  AM  I F  and  AN  /  Q  are  not  proper  fractions,  reduce  them 
to  sums  of  integral  functions  and  proper  fractions,  and  let  the 
results  be 

4  If  R  A  N  (' 

(2) 


AM 

B 

AN 

r 

■  1  + 

and 

■:    A    + 

P 

P 

Q 

Q 

Substituting  these  expressions  for  AM  J  P  and  AN  /  Q  in  (1), 
we  have 

But  since  A/PQ,  B/P,  and  C /Q  are  proper  fractions,  it 
follows  from  (3),  by  §§  530,  531,  that  7  +  A'  =  0  and 
A    _B       C 
PQ~  P       q' 
as  was  to  be  demonstrated. 

533  Note.     The   fraction  A/PQ  can   be  reduced  to  but  one  such  sum 
B/P+C/Q. 

A        B      C      B'      C 
lor  suppose  = —  = \-^i  (1) 

PQ    p    q    p    q  ^  ' 

where  B'  /P  and  C  /  Q  also  denote  proper  fractious. 

Then  ^^^  =  ■^-^,  and  therefore  ^^  ~  ^'^  ^  =C'-C.  (2) 

But  (2)  is  impossible  unless  B  —  E'  =  0  and  C  —  C  =  0.  For  other- 
wise (2)  would  mean  that  (B  —  B')  Q  is  exactly  divisible  by  P,  and  this 
cannot  be  the  case  since  Q  is  prime  to  P  and  B  —  B'  is  of  lower  degree 
than  P,  §  481. 

534  Partial  fractions.    We  call  the  fractions  B/P  and  C/Q,  whose 
existence  we  have  just  T^xovedi,  partial  fractions  of  A/PQ. 

To  resolve  a  given  fraction  of  the  form  A  / PQ  into  its  par- 
tial fractions  B/P  and  C / Q,  it  is  not  necessary  to  carry  ouL 
the  process  indicated  in  §  532 ;  we  may  apply  the  method  of 
undetermined  coefficients,  §  397,  as  in  the  following  example. 

Example  1.  Resolve  {2 x"^  +  \)  / {z^  —  V)  into  a  sum  of  two  partial 
fractions. 

This  is  a  proper  fraction,  and  its  denominator  is  a  product  of  two 
factors,  X  —  1  and  x^  +  x  +  1,  which  are  prime  to  each  other. 


RATIONAL    FRACTIONS  239 

Hence,  §  532,  (2x2  +  l)/(x^  —  1)  is  equal  to  a  sum  of  two  proper  frac- 
tions whose  denominators  are  a:  —  1  and  x^  +  x  +  1  respectively.     The 
numerator  of  the  first  of  these  fractions  must  be  a  constant,  that  of  the 
second  an  expression  whose  degree  is  one  at  most.     Hence  we  must  have 
2x2+l_     a  bx  +  c 

x^-1   ""  x-1  "^  x2  +  X  +  1  ^^^ 

where  a,  6,  and  c  denote  constants. 

To  find  a,  b,  c,  clear  (1)  of  fractions. 

We  obtain    2  x^  +  1  =  a  (x-  +  x  +  1)  +  (6x  +  c)  (x  -  1), 
or  2x2  +  1  =  (a  +  6)x2  +  (a  -  6  +  c)x  +  (a  -  c).  (2) 

As  (2)  is  an  identity,  the  coefficients  of  like  powers  of  x  are  equal,  §  284. 

Hence  a  +  b-2,  a-b  +  c  =  0,  a-c  =  l,  (3) 

or,  solving  (3),  a  —  I,  6  =  1,  c  =  0. 

Therefore     ^^!±i  ^  ^A^  + ? '. 

X3  -  1  x-1         X2  +  X  +  1 

Example  2.  Resolve  (5x  +  4)/(x'*  +  x^  +  x^  —  x)  into  a  sum  of  two 
partial  fractions. 

General  theorem  regarding  partial  fractions.     From  the  theorem     535 
of  §  532  we  may  draw  the  following  conclusions. 

1.  Let  A  / PQR  denote  a  proper  fraction  in  which  the  three 
factors  of  the  denominator  P,  Q,  R  are  prime  to  one  another. 
This  fraction  can  be  reduced  to  a  sum  of  the  form 

A     _B      D      E 
PQR  ~  P       Q      R 
where  B  fP,  D /  Q,  and  E  /  R  denote  proper  fractions. 

For  since  P  is  prime  to  QR,  §  482,  A  / PQR  is  the  sum  of 
two  proper  fractions  of  the  form  B/P+  C /  QR,  §532;  and 
since  Q  is  prime  to  R,  C  J  QR  is  itself  the  sum  of  two  proper 
fractions  of  the  form  D/Q  +  E/R,%  532. 

The  like  is  true  when  the  denominator  is  the  product  of 
any  number  of  factors  all  prime  to  one  another. 

2.  Consider  the  proper  fraction  A  / PQ^  in  which  P  is  prime 
to  Q.     By  §  532  it  can  be  resolved  into  the  sum 

A    _B       C 
PQ^~  P      q^' 


240  A   COLLEGE   ALGEBRA 

We  cannot  apply  the  theorem  of  §  532  to  the  fraction  C I  Cl\ 
since  the  factors  Q,  Q,  Q  are  not  prime  to  one  another. 

But  since  C  is  of  lower  degree  than  Q^  it  can  be  reduced, 
§  422,  to  a  polynomial  in  Q  of  the  form 

C  =  CiQ2  +  C2Q  +  C3, 
where  Ci,  C2,  and  Tg  are  of  lower  degree  than  Q. 

Dividing  each  member  of  this  identity  by  Q^  we  have 

Hence  the  given  fraction  can  be  reduced  to  the  sum 
A    _  ^       Ci       C'2       C3 

where  B  is  of  lower  degree  than  P,  and  C^,  C^,  C3  are  of  lower 
degree  than  Q. 

And  so  in  general  when  a  factor,  as  Q,  occurs  more  than 
once  in  the  denominator. 

We  therefore  have  the  following  theorem. 

Suppose  that  the  denominator  of  a  given  proper  fraction  has 
been  separated  into  factors  —  some  occurring  once,  some,  it  may 
he,  more  than  once  —  which  are  all  prime  to  one  another. 

The  fraction  itself  can  then  be  resolved  into  one  and  but  one 
sum  of  proper  fractions  in  which  (1)  for  each  factor,  P,  which 
occurs  but  once,  there  is  a  single  fraction  of  the  form  B/P,  and 
(2)  for  each  factor,  Q,  tvhich  occurs  r  times,  there  is  a  group 

of  r  fractions  of  the  form  Ci/Q  +  Cj/Q'^  H h  C,/Q^  ivhere 

Ci,  C2,  •  •  •,  C,  are  all  of  lower  degree  than  Q. 

536  Simplest  partial  fractions.  It  can  be  proved  that  ever}^  poly- 
nomial in  X  with  real  coefficients  is  the  product  of  factors  of 
one  or  both  the  types  x  —  a  and  a;^  +  px  +  q,  where  a,  p,  and 
q  are  real,  but  where  the  factors  of  x"^  +  />x  +  q  have  imagi- 
nary coefficients. 

Moreover  it  follows  from  §§  469.  532  that,  if  the  numerator 
of  a  given  proper  fraction   and  the   factors   into  which  its 


RATIONAL    FRACTIONS  241 

denominator  has  been  separated  have  real  coefficients,  so  will 
the  numerators  of  the  corresponding  partial  fractions.  Hence, 
by  §  534, 

Every  'proper  fraction  whose  mimerator  and  denominator 
have  real  coefficients  is  equal  to  a  definite  sum  of  partial  frac- 
tions related  us  follows  to  the  factors  x  —  a  and  x^  +  px  +  q  o/" 
its  denominator. 

1.  For  every  factor  x  —  a  occurring  once  there  is  a  single 
fraction  of  the  form  A/(x  —  a),  xoliere  A  is  a  real  constant. 

2.  For  every  factor  x  —  a  occurring  r  times  there  is  a  group 
of  r  fractions  of  the  form 

Ai/(x  -  a)  +  k,/{x  _  a)2  +  •  ■  ■  +  A,/(x  -  a)^ 
where  Aj,  Ag,  •  •  •  A^  are  real  co?istanfs. 

3.  For  every  factor  x^  -|-  px  +  q  occurring  once  there  ts  a, 
single  fraction  of  the  form  (Dx  +  E)/(x'^  -f-  px  +  q),  ivhere  D 
and  E  are  real  constants. 

4.  For  every  factor  x^  +  px  +  q  occurring  r  times  there  is 
a  group  of  r  fractions  of  the  form 

(D,x  4-  E0/(x2  +  px  +  q)  +  •  •  •  +  (D,x  +  E,)/(x2  +  px  +  q)^ 
where  Dj,  Ej,  Do,  Ej,  •  •  •  D^,  E^  denote  real  co)istants. 

The  fractions  here  described  are  usually  called  the  simplest     537 
partial  fractions  of  the  given  fraction.     They  are  best  found 
by  the  method  of  undetermined  coefficients. 

ifl  -\-  X  —  3 
Example  1.     Resolve  into  its  simplest  partial 

fractions. 

By  §  536,  we  have ^^^ =  -^  +  -^  +  -^—    (1) 

(X  -  1)  (X  -  2)  (x  -  3)      X  -  1      X  -  2      X  -  3    ^  ' 

where  A,  B,  C  are  constants. 

Clearing  (1)  of  fractions,  we  obtain 
x2  +  X  -  3  =  4  (X  -  2)  (x  -  3)  +  B(x  -  3)  (X  -  1)  +  C(x  -  1)  (X  -  2).     (2) 

We  may  find  A,  B,  C  by  arranging  the  second  member  of  (2)  accord- 
ing to  powers  of  x  and  equating  coefficients  of  like  powers  ;  but,  since 
A,  B,  C  are  constants,  the  same  results  will  be  obtained  by  the  following 
method,  which  is  simpler. 


242  A   COLLEGE    ALGEBRA 

In  (2)  let  a;  =  1,  and  we  have  -  1  =  ^  ( -  1)  ( -  2),   .-.A  =-1/2 
next         let  x  =  2,  and  we  have      3  =  5  ( -  1)  •  1,        .-.  B  =  -  H; 
finally      let  a:  =  3,  and  we  have      9=C-2-l,  .-.(7  =  9/2. 

Hence  = h 


(X  -  1)  (I  -  2)  (x  -  3)  2  (X  -  1)      X  -  2      2  (X  -  3) 

X  +  1 

Example  2.     Resolve into  its  simplest  partial  fractions 

x(x-lf 

X,     o  ro^          ,.             a;  +  l  A  B  C  B 

By  §  536,  we  have =  —  + 


x(x-l)3       X       x-1       (x-l)^      {x-lf 
and  therefore     x  +  1  =  J.  (x  -  1)''  +  J5x  (x  -  1)-  +  Cx  (x  -  1)  +  Dx.     (1) 

In  (1)  let  X  =  0,  and  we  have  1  =  J.  (-  1)3,  .■.A=-l; 
next         let  x  =  1,  and  we  have  2  =  J),  .-.  D  =  2. 

Substitute  these  values  of  A  and  D  in  (1),  transpose  to  the  first  mem- 
ber the  terms  thus  found,  namely  —  (x  —  1)^  and  2x,  and  simplify  the 
result.     We  obtain 

x3  -  3  x2  +  2  X  =  5x  (X  -  1)-  +  Cx  (X  -  1).  (2) 

Dividing  both  members  of  (2)  by  x(x  —  1),  we  have 

x-2  =  B(x-l)+C.  (3) 

Equating  coefiicients  of  like  powers  of  x  in  (3),  we  have 

1  =  B  and  -  2  =  -  B  +  C,    .-.  B  =  1  and  C  =  -  1. 

x+1              11                1,2 
Hence        = 1 \- 


x(x-l)3         X      x-1      (x-l)2      (X  -  1)3 
Or  we  may  arrange  (1)  according  to  powers  of  x,  obtaining 
X  +  1  =  {A  +  B)x^  ~  {S  A  +  2  B  -  C)x^  +  {3  A  +  B  -  C  +  D)x  -  A. 
Equating  coeflQcients  of  like  powers  of  x,  we  have 
^  +  -5  =  0,  3A  +  2B-C  =  0,  3A  +  B  -  C  +  D  =  -[,  -  A  =  I. 
And  from  these  equations  we  find,  as  before, 

A=-l,  B  =  l,  C  =  -  1,  D  =  2. 

5  a;2  —  4  X  -I-  16 

Example  3.      Resolve  — — —- into   its  simplest  partial 

\X   —  X  +  1 )  -  (x  —  3) 
fractions. 

The  factors  of  x^  -  x  +  1  being  imaginary,  we  have,  §  536, 

5x2  -4x  -f  16       ^      Ax  +  B  Cx  +  D  E 

(X2  -  X  +  1)'^  (X  -  3)  ~  (X2  -  X  +  1)2         X2  -  X  +  1  "*"  X-S' 

where  A,  B,  C,  D,  E  are  constants. 


RATIONAL   FRACTIONS  243 

Clearing  of  fractions, 

6x2 -4x  +  W^{Ax  +  B){x-S) 

+  {Cx  +  D){x^-x  +  l){x~3)  +  E{x^-x  +  1)2.     (1) 

"We  may  find  A,  B,  C,  B,  E  by  arranging  (1)  according  to  po^Yers  of 
X  and  then  equating  coefficients  of  like  powers  ;  but  the  following  method 
is  simpler. 

In  (1)  let  X  =  3,  and  we  have  49  -  49^,  .-.  E  ~\. 

Substitute  this  value  of  E  in  (1),  transpose  the  term  (x^  —  x  +  1)2 
thus  found  to  the  first  member,  simplify,  and  divide  both  members 
b^  X  -  3.     We  obtain 

-  (x3  +  x2  +  X  +  5)  =  ^x  +  i?  +  (Cx  +  D)  (x2  -  X  +  1).  (2) 

Next  divide  both  members  of  (2)  by  x2  —  x  +  1.     We  obtain 

2x  +  3           Ax  +  B 
-  X  -  2  - ^— --  = ^—  +  CX  +  I).  (3) 

X2-X+Ix2-X  +  1  ^ 

By  §  531,  the  fractional  parts  and  the  integral  parts  in  (3)  are  equal. 

Hence     -  x  -  2  =  Cx  +  D,  and  therefore  C  =  -  1,  Z)  =  -  2, 

and  -2x  —  3=J.x+  B,  and  therefore  J.  =  -  2,  j5  =  -  3. 

^,       ^  5x2-4x  +  16  2x  +  3  x  +  2  1 

Therefore  = 1 

(x2  -  x  +  1)2  (X  -  3)  (x2  -  X  +  1)2      a;2  -  X  +  1      X  -  3 

When  the  denominator  of  the  given  fraction  has  the  form 
(x  —  ay,  it  is  best  to  begin  by  expressing  the  numerator  in 
powers  of  a;  —  a,  §  423.  Similarly  when  the  denominator  has 
the  form  {x^  +  px  +  qY,  the  factors  of  x"^  +  px  +  «/  being  imagi- 
nary, we  express  the  numerator  in  powers  of  x"  +  p^  +  ?• 

X*  4-  x3  _  8  x2  +  6  X  32 

Example.     Resolve  into  its  simplest  partial 

fractions. 

By  §  423,  we  find 
x*  +  x3  -  8  x2  +  6 X  -  32  =  (X  -  2)*  +  9(x  -  2)3  +  22(x  -  2)2  +  18 (x  -  2)  - 28. 

Dividing  both  members  by  (x  —  2)^,  we  have 

x^  +  x3-8x2  +  6x-32_      1  9  22  18 28_ 

(x-2)5  ~x-2       (x-2)2      (x-2)3       (x-2)*      {x-2)5" 

If  given  an  improper  fraction,  we  may  first  reduce  it  to  the 
sum  of  an  integral  expression  and  a  proper  fraction  and  then 
resolve  the  latter  into  its  partial  fractions. 


244  A   COLLEGE   ALGEBRA 

x3- 2x2 -6a; -21 


Example.     Apply  this  method  to  the  fraction 


x-!  —  4  X  —  6 


^.    ,           x3-2x2-6x-21  „  7x-ll 

We  have    =  x  +  2  + 


x2_4x-5  x2-4x-5 

7x-ll 


X  +  2 


(X  +  1)  (X  -  5) 
^nd  proceedlDg  as  above  we  find 

7x-ll  3  4 


(X  +  1)  (X  -  5)      X  +  1      X  -  5 


EXERCISE   XXVm 


Resolve  the  following  into  the  simplest  partial  fractions  whose  denomi- 
nators have  real  coefficients. 


1 

2x+  11 

2. 

6x-l 

(X  -  2)  (X  +  3) 

(2x  +  l)(3x  -  1) 

3 

4x 

4 

x2  +  2  X  +  3 

(X  +  1)  (X  +  2)  (X  +  3) 

{x-l){x-2)(x-3)(x-4) 

5. 

x2  +  2 

1+X3 

6. 

8x  +  2 

X-X3 

7 

x^-x^-~5x\-i 

8 

2X3-X2  +  1 

x2-3x  +  2 

(X  -  2)* 

9 

x-1 

10. 

6 

2x3-5x2-12x 

2  X*  -  x2  -  1 

n 

2x3-3x2  +  4x-5 

12 

X2  +  X  +  1 

(x  +  3)s 

(.C2+   l)(x2  +  2) 

n 

x2  +  6  X  -  1 

14 

3x  -1 

(X-  3)2  (x-1) 

(X  -  2)  (X2  +  1) 

1*) 

2  x6  -  X  +  1 

16 

2x2 -x  +  1 

(X2  +  X+  1)3 

(X2  -  X)2 

17. 

3x2- X +  2 
(x2  +  2)  (X2  -  X  -  2) 

18. 

x^+px  +  q 
(x  -  a)  (X  -  b)  (X  -  c) 

19. 

2x2 -3x  -2 

20. 

X3  +  X  +  3 

X(X-l)2(x  +  3)3  X*+X2+1 


SYMMETRIC    FUNCTIONS  245 

IX.     SYMMETRIC    FUNCTIONS 

ABSOLUTE  SYMMETRY   AND  CYCLO-SYMMETRY 

Absolute  symmetry.  In  the  expression  x'^  -\-  y^  -{-  z^  the  let-  540 
ters  a-,  y,  z  are  involved  in  such  a  manner  that  if  any  two  of 
them  be  interchanged  a;^  +  ^/^  +  z^  is  transformed  into  an  iden- 
tically equal  expression,  namely,  y'^  -\-  x^  -\-  z"^,  or  z"^  -\-  y"^  +  x^, 
or  x^  -{-  z^  +  \f.  To  indicate  this,  we  say  that  x"^  -\-  y"^  ■\-  z^  is 
symvietric  with  respect  to  x,  y,  z. 

And,  in  general,  a  function  of  a  certain  set  of  letters  is  said 
to  be  symmetric  with  respect  to  these  letters  when  every  inter- 
change of  two  of  the  letters  will  transform  the  function  into 
an  identically  equal  functioi-. 

Other  examples  of  symmetric  functions  are 

(xy  +  xz  +  yz)/{x  +  y)  (x  +  z)  {y  +  z)  with  respect  to  a;,  y,  z, 
a-\-h  -\-  c  and  (x  +  a)  (x  +  h)  (x  +  c)  with  respect  to  a,  6,  c. 

On  the  other  hand,  x  -|-  ?/  —  z  is  not  symmetric  ;  for  if  we  interchange 
y  and  z  we  obtain  x  +  z  —  y,  which  is  not  equal  to  x  +  ?/  —  z. 

We  call  2  xhj  and  3  yH  terms  of  the  savie  type  with  respect     541 
to  the  variables  cc,  ?/,  z,  because  the  variable  parts  of  these  terms, 
namely,  x^y  and  y'^z,  can  be  transformed  into  one  another  by 
interchanges  of  pairs  of  the  letters  x,  y,  z.     And  so  in  general. 

The  sufficient  ayid  necessary  condition  that  an  integral  func-     542 
tion  of  certain  letters,  as  x,  y,  z,  he  symmetric  with  respect  to 
these  letters  is  that  all  its  terms  of  the  same  type  shall  have  the 
same  coefficie7its. 

This  implies  that  if  a  symmetric  function  contains  one  term 
of  a  certain  type,  it  must  contain  all  terms  of  that  type ;  that 
is,  all  terms  that  can  be  derived  from  the  term  in  question  by 
making  every  jwssible  interchange  of  the  letters. 

Thus,  if  ax-  +  by-  +  cz"^  is  to  be  symmetric,  we  must  have  a  =  b  =  c. 
Again,  if  a  symmetric  function  of  x.  ?/,  z  contains  the  term  x'^y,  it 
must  contain  all  the  terms  x'^y  -(-  y-x  4-  x^z  +  z^x  -f  y'^z  +  z'^y. 


246  A   COLLEGE    ALGEBRA 

543  This  theorem  will  indicate  the  general  form  of  a  symmetric 
function  of  given  degree  with  respect  to  a  given  set  of  letters. 

Thus,  the  general  form  of  a  symmetric  function  of  the  first  degree 
with  respect  to  x,y,  z,  u  i&  a{x  +  y  ^  z  -[■  u)  +  h,  where  a  and  6  denote 
constants. 

Again,  the  most  general  symmetric  and  homogeneous  functions  of  the 
first,  second,  and  third  degrees  with  respect  to  x,  y,  z  are 

1.    a{x  +  y+  z).  2.    a  (x2  +  y'i  +  z^)-\-h  {xy  +  xz  +  yz). 

3.    a  (x3  +  2/3  +  2-5)  +  b  (xhj  +  y'-x  +  x^z  +  z'^x  +  y^z  +  z'^y)  +  cxyz. 

544  On  expressing  a  symmetric  function.  The  notation  'Zx"  means 
the  sum  of  all  terms  of  the  same  type  as  x"^ ;  that  is,  if  a-,  y,  z  are 
the  letters  under  consideration,  %x-  =  x'^  +  if  +  z^.  Similarly 
^x^y  =  xhj  +  y'^x  +  x-'z  +  z^x  +  y^z  +  zhj ;  and  so  on. 

Any  given  symmetric  function  may  be  represented  by  select- 
ing from  its  terms  one  of  each  type,  and  writing  the  symbol  2 
before  their  sum. 

Thus, 
S  (2  X  -  x3?/2)  =  2x  +  2y  +  2z  -  x^y^  -  t/'^x-  -  x^z^  -  z^x'^  -  y^z^  -  z^y^. 

545  When  writing  out  symmetric  functions  at  length,  it  is  best 
to  arrange  the  terms  in  accordance  with  some  fixed  rule.  The 
following  examples  Avill  indicate  a  convenient  rule  for  the 
arrangement  of  integral  symmetric  functions. 

Suppose  that  the  letters  under  consideration  are  a,  6,  e,  d,  and  by  the 
normal  order  of  these  letters  understand  the  order  a,  6,  c,  d. 

We  shall  then  write  "Lab  and  Ea6c  as  follows  : 

2a6  =  a6  +  ac  +  ad  +  6c  +  hd  +  cd,  I,abc  =  abc  +  abd  +  acd  +  bed. 

Observe  that  in  each  term  we  write  the  letters  in  their  normal  order. 

In  forming  Hah  we  take  each  letter  a,  6,  c  in  turn  and  after  it  write  each 
subsequent  letter.  The  terms  of  Zabc  are  derived  in  a  similar  manner 
from  those  of  2a&. 

We  shall  arrange  the  terms  of  Sa'"6",  when  m  ^f  n,  as  follows : 
Sa263  =  a253  +  62^3  +  a^c^  +  c^a^  +  •  •  •  +  c^d^  +  d'^c'^. 

Observe  that  we  keep  the  order  of  the  exponents  fixed  ;  then  under  the 
exponents  we  write  the  letters  of  the  first  term  of  'Lab  in  both  the  orders 
ab  and  ba,  and  so  on. 


SYMMETRIC    FUNCTIONS  247 

In  like  manner  we  may  write 

Sa263c4  =  a263c4  +  a'^c^b*  +  b^-c^a^  +  b^a^c*  +  c^a^b*  +  c'^b^a* 
+  (terms  similarly  derived  from  the  remaining  terms  of  'Labc). 

A  general  theorem  regarding  symmetry.     It  follows  from  the     546 
definition  of  symmetry,  §  540,  tliat  a  symmetric  function  will 
remain  symmetric  when  its  form  is  changed  by  the  rules  of 
reckoning.     In  particular, 

The  sum,  difference,  jjroduct,  and  quotient  of  two  symmetric 
functions  are  themselves  symmetric. 

By  aid  of  this  theorem  we  may  obtain  the  result  of  combin- 
ing given  symmetric  functions  by  algebraic  operations  without 
actually  carrying  out  these  operations.  It  is  only  necessary  to 
compute  the  various  typical  terms  of  the  result. 

Example  1.     Find  (Sa)-  =  (a  +  b  -{■  c  -^ )2. 

Evidently  the  required  result  is  a  homogeneous  symmetric  function  of 
the  second  degree  consisting  of  terms  of  the  two  types  cfi  and  2  ab. 
Hence  (Sa)2  =  2a2  +  2  Sa6. 

Example  2.     Find  2x2  •  Sx  =  (x2  ^  yi  j^  z^){x  +  y  +  z). 

Evidently  this  product  is  a  sum  of  terms  of  the  two  types  x^  and  x'^y. 

Hence       2x2  .  sx  =  2x3  +  2x2?/  ^x^  +  y^  +  z^ 

+  x^y  +  2/2x  +  x^-z  +  z^x  +  y-z  +  z^y. 

Example  3.     Find  (2x)3  =  {x  +  y  +zf. 

The  required   result  is  homogeneous,   symmetric,   and  of   the   third 
degree  with  respect  to  x,  y,  z.     We  must  therefore  have,  §  543, 
{x  +  y  +  zY  =  a  (x3  +  2/3  ^  23)  +  6  {x^y  +  yH  +  x2z  +  z2x  +  y'^z  +  z'^y)  +  cxyz. 

To  find  the  values  of  the  constants  a,  b,  c,  assign  any  three  sets  of 
values  to  x,  y,  z  which  will  yield  equations  in  a,  6,  c,  and  solve  these 
equations. 

Thus,  putting      x  =  1,  2/  =  0,  z  =  0,  we  have    1  =  a.  (1) 

Again,  putting    x  =  1.  ?/  =  1,  z  =  0,  we  have    8  =  2  a  +  26.  (2) 

Finally,  putting  x  =  1,  2/  =  1,  2  =  1,  we  have  27  =  3o  +  6&  +  c.     (3) 

Solving  (1),  (2),  (3),  we  obtain  a  =  1,  6  =  3.  c  =  6. 

Hence  (2x)3  =  2x^  +  3  2x2?/  +  6  2x2/z. 


248  A    COLLEGE    ALGEBRA 

547  Cyclo-symmetry.  In  the  expression  x^y  +  y"z  +  z^x  the  let- 
ters X,  y,  z  are  involved  in  such  a  manner  that  if  we  replace  x 
by  y,  y  by  z,  and  z  by  x,  we  obtain  an  identically  equal  expres- 
sion, namely,  y'^z  +  z^x  +  x-y.  To  indicate  this,  we  say  that 
x^y  +  y'^z  +  z^x  is  cyclo-symmetric,  or  cyciJie,  with  respect  to  the 
letters  x,  y,  z,  taken  in  the  order  x,  y,  z. 

And,  in  general,  an  expression  is  said  to  be  cyclo-symmetric, 
or  cyclic,  with  respect  to  certain  letters  arranyed  in  a  given 
order,  if  it  is  transformed  into  an  identically  equal  expression 
when  we  replace  the  first  of  these  letters  by  the  second,  the 
second  by  the  third,  and  so  on,  and  the  last  by  the  first. 

Such  an  interchange  of  the  letters  is  called  a  cyclic  interchange. 

548  Observe  that  the  terms  of  x'^y  +  y'^z  +  z^x  are  themselves 
arranged  cyclicly ;  that  is,  so  that  the  first  changes  into  the 
second,  the  second  into  the  third,  and  the  third  into  the  first, 
when  we  replace  x  hy  y,  y  by  z,  and  z  by  x.  Cyclic  expres- 
sions are  of  frequent  occurrence  and  reckoning  with  them  is 
greatly  facilitated  by  arranging  them  cyclicly. 

549  Evidently  every  symmetric  function  is  cyclic,  but  not  every 
cyclic  expression  is  symmetric. 

Thus,  x'^y  -\-  y'^z  +  z-x,  though  cyclic,  is  not  symmetric.  Its  vahie 
changes  if  x  and  y  are  interchanged.  To  make  it  symmetric  we  must 
add  the  group  of  terms  y'^x  -\-  z'^y  -f-  x^z. 

550  As  the  example  shows,  a  cyclic  function  will  ordinarily  not 
contain  all  the  terms  of  a  given  type,  but  such  of  these  terms 
as  it  does  contain  will  have  the  same  coefficients. 

551  Theorem.  The  sum,  difference,  product,  and  quotient  of  tivo 
cyclic  fujictions  are  themselves  cyclic. 

This  follows  at  once  from  the  definition  of  cyclo-symmetry. 
Example.     Find  the  product  {x^y  +  yH  -f  z'^x)  (x  -t-  ?/  -|-  z). 
Evidently  the  product  is  cyclic  but  not  symmetric.     Moreover  it  con- 
tains the  terms  x^y,  x'^y-,  x"yz,  each  once,  and  terms  of  these  types  only. 
Hence  the  product  is 

x^y  +  yH  Jr  z^x  +  x-y-  +  yH-  -f  zH"^  -\-  x'^yz  ■\-  yHx  -\-  zHy. 


SYMMETRIC    FUNCTIONS  249 

EXERCISE  XXIX 

1.  State  the  letters  with  respect  to  which  the  expression 

X4  -  2  2/*  +  2^  +  4  (X3  -  2/3)  (2/3  -  23)  (x^  +  ^2) 

is  symmetric. 

2.  Write  out  in  full  the  following  symmetric  functions  of  a,  6,  c. 
Sa262,         sa364,         2aV&,         2a263c5,         Sa^ft'^c*, 

S(a  +  6)c,  S(a  +  62)c3,  S(a  +  26  +  3c). 

3.  Show  that  (a  —  b)  (b  —  c)  (c  —  a)  is  cyclic  but  not  symmetric  with 
respect  to  a,  b,  c  ;  also  that  (a  —  b)^{b  —  c)^{c  —  a)^  is  symmetric. 

4.  Is  (a  —  6)2  (6  —  c)2  (c  —  d)'^  (d  —  a)^  symmetric  with  respect  to 
a,  6,  c,  d? 

5.  Arrange  the  following  sets  of  expressions  cyclicly. 

y-  —  X",  2^  —  2/2,  x~  —  2- ;     a-bc,  abd'^,  acH,  b'^cd  ; 
{a  -  c)  (6  -  a),  (a  -c)(c-  b),  {a  -b)(b-  c). 

6.  Write  out  in  full  the  cyclic  functions  of  a,  6,  c,  d  whose  first 
terms  are 

a63c2,       a{b-  c),       (b  +  2c){a  +  d),       a"- /{a  -b){a-  c). 

7.  Prove  the  truth  of  the  following  identities. 

Sa3  •  Sa  =  2a*  +  ^a^b  ;     Zab  ■  2a  =  2a-'6  +  3  2a6c. 

FACTORIZATION   OF   SYMMETRIC  AND   CYCLIC   FUNCTIONS 

By  aid  of  the  remainder  theorem  and  the  principles  just     552 
explained  it  is  often  possible  to  factor  a  complicated  symmetric 
or  C3^clic  function  with  comparatively  little  reckoning. 

Example  1.     Factor  x^  {y  —  z)  +  y^  {z  —  x)  -\-  z^  (x  —  y). 
This  function  vanishes  when  y  =  z^  for 

X3  (2  -  2)   +  23  (2  -  X)   +  z3  (x  -  2)  =  0. 

Hence  the  function  is  exactly  divisible  by  ?/  —  2,  §  416  ;  and  for  a  like 
reason  it  is  exactly  divisible  by  z  —  x  and  by  x  —  2/,  and  therefore  by  the 
product  (2/  —  z){z  —  x)  (x  —  y). 

Both  dividend  and  divisor  are  cyclic  and  homog»neous,  and  their 
degrees  are  four  and  three  respectively.     Hence  the  quotient  must  be 


250  A   COLLEGE   ALGEBRA 

a  cyclic  and  homogeneous  function  of  the  first  degree  and  therefore  have 
the  form  k(x-i-  y  +  z),  where  k  denotes  some  constant.     Hence 
x^  {y-z)+  2/3  (z-x)+z-{x-i/)~k{y-  z)  {z -x){x  ~y,{x  +  y  ^-  z)    (1) 
To  find  k,  assign  to  x,  y,  z  any  set  of  values  for  which  the  coefficient 
of  k  will  not  vanish. 

Thus,  putting  x  =  2,  y  =  1,  z  =  0,  we  have  6  =  —  6A;,  orA:  =  — L 
Or  we  may  find  k  by  equating  the  coefficients  of  like  powers  of  x 
in  the  two  members  of  (1)  arranged  as  polynomials  in  x.     Thus,  the  x^ 
term  in  the  first  member  is  x^  {y  —  z),  and  in  the  second  member  it  is 
—  kx^{y  —  z),  whence  as  before  fc  =  —  1.     We  therefore  have 

X3  {y  -Z)  +  2/3  (Z  -  X)  +  Z3  (X  _  y)  =  _  (y  -  z)  (z  -  X)  (X  -  ?/)  (x  +  ?/  +  z). 

Example  2.     Factor  {x  +  y  +  zY  —  x^  -  y^  —  z^. 

This  function  vanishes  when  x  =  —  y,  for 

{-  y  +  y  +  zY  +  y^  -  y^  -z^  =  0. 

Hence  the  function  is  exactly  divisible  by  x  +  2/ ;  and  for  a  like  reason  it 
is  divisible  by  2/  +  z  and  by  z  +  x,  and  therefore  by  (x  +  y)  {y  +  z)  (z  +  x). 

As  the  dividend  and  divisor  are  symmetric  and  homogeneous  and  of 
the  fifth  and  third  degrees  respectively,  the  quotient  must  be  of  the  form 
k  {x2  +  2/2  +  z2)  +  i  (xy  +  yz-\-  zx),  §  543. 

Hence  {x  +  y  +  zf  -  x^  -  y^  -  z^ 

=  (x-Vy){y  +  z)  (z  +  X)  [k  (x^  -f  y'^  +  z'-^)  +  Z  (X2/  +  2/2  +  zx)]. 

Putting       X  =  1,  2/  =  1)  2  =  0,  we  obtain  15  =  2  fc  +  L 

Putting       X  =  2,  2/  =  1)  z  =  0,  we  obtain  35  =  5  A:  +  2  i. 

Solving  for  A;  and  i,  we  find  fc  =  5,  Z  =  5,  and  therefore  have 
(x-\-y  +  z)^  -x^  -y^  -  z5 

=  5(x  +  2/)  (2/  +  z)  (z  +  X)  (x2  +  2/2  +  z2  +  a;y  +  2/z  +  zx). 

Example  3.     Factor 

{x  +  y  -\-zY  -{y  -\-  z-xY  -  {z  +  x-yY  -  {x  +y  -  zY- 

This  function  vanishes  when  x  =  0,  for 

(2/  +  zY  -  (2/  +  z)3  -  (z  -  yY  -(y-  zY  =  0. 

Hence  the  function  is  exactly  divisible  by  x  -  0  or  x ;  and  for  a  like 
reason  it  is  divisible  by  y  and  by  z  and  therefore  by  xyz. 

Since  both  dividend  and  divisor  are  of  the  third  degree,  the  quotient  is 
some  constant,  k.     Hence 

(x  +  y  +  zY-{y  +  z-xY-(z  +  x-yY-ix  +  y-  zY  =  kxyz. 
Putting  x  =  1, 2/  =  1,  z  =  1,  we  find  k  =  24,  and  therefore  have 

{x  +  y  +  zY-{y  +  z-xY-{z  +  x-yY-{x  +  y-zY  =  2ixyz. 


SYMMETRIC    FUNCTIONS  251 

The  method  just  explained  is  sometimes  useful  in  simplifying     553 
cyclic  fractional  expressions. 

a^  h^  c3 

Example.     Simplify 1- 


{a-h){a-c)      {b-c)(b-a)      (c-a){c-b) 

This  expression  is  cyclic  with  respect  to  a,  6,  c. 

The  lowest  common  denominator  is  [b  —  c)  (c  —  a){a  —  b). 

On  reducing  the  fractions  to  this  lowest  common  denominator  we 
obtain  as  the  first  numerator  —  a^  (b  —  c).  Hence,  by  cyclo-symmetry, 
the  second  and  third  numerators  are  —b^(c  —  a)  and   —  c^{a  —  b). 

Adding  these  three  numerators  and  factoring  the  result,  §  552,  Ex.  1, 
we  have  (ct  +  &  +  c)  (6  —  c)  (c  —  a)  {a  —  b). 

Hence  the  given  expression  reduces  to  a  +  6  +  c. 

EXERCISE  XXX 

Factor  the  following  expressions. 

1.  x2(y  -  2)  +  r/-2  (z  -  x)  +  z'^  (X  -  y). 

2.  yz  {y  -  z)  +  zx  {z  -  x)  +  xy  (x  -  y). 

3.  {y  -  zY  +  {z-  x)3  +  (X  -  y)\ 

4.  x{y  -  z)3  +  ^  (z  -  x)3  +  z  (x  -  ?/)3. 
6.    x2  (y  -  zf  +  ?/  (z  -  x)3  +  z2  (x  -  y)\ 

6.  X*  (2/2  -  22)  +  2/«  (Z2  -  X2)  +  Z*  {x2  -  2/2). 

7.  (x  +  2/  +  2)3  —  x3  —  2/3  —  2^. 

8.  {y  -  2)6  +  (2  -  x)5  +  (X  -  2/)5. 

9.  (x  +  2/  +  z)5  -  (2/  +  2  -  x)5  -  (z  +  X  -  2/)^  -  (X  +  2/  -  zf. 

10.  (2/  -  2)  (2/  +  2)3  +  (z  _  X)  (2  +  X)3  +  (X  -  2/)  (X  +  2/)"- 

11.  x{y  +  2)2  +  2/(z  +  X)2  +  2(X  +  2/)2  -  4X2/2. 

12.  X5  (2/  -  2)  +  2/^  (Z  -  X)  +  Z5  (X  -  ?/). 

Simplify  the  following  fractional  expressions. 
a*  b*  c* 


13 


14. 


(a  -  6)  (a  -  c)      (6  -  c)  (6  -  a)      (c  -  a)  (c  -  b) 

X  +  a  X  +  b  x  +  c 

^a"^^^6Ha^^      (6  -  c)  (6  -  a)      (c  -  a)  (c  -  6)  ° 


252  A   COLLEGE   ALGEBRA 


a*  —  6c  b"  —  ca  c^  —  ab 

16.    : — -  + 


(a  -  6)  (a  -  c)      (b  -c)(b  -  a)      (c  -  a){c  -  b) 
16.         (&  +  c)2         ^        jc  +  a)^        ^        (a +  6)2 


17. 


(o  -  6)  (a  -  c)      (6  -  c)  (6  -  a)      (c  -  a)  (c  -  6) 

a2  62  c2 


(a  -  6)  (a  -  c)  (X  -  a)      (6  -  c)  (6  -a){x-  b)      (c  -  a)  (c  -  6)  (x  -  c) 


X.     THE    BINOMIAL   THEOREM 

654         Structure  of  continued  products.     To  obtain  the  product 

(a  +  b  i-  c  +  d)(e  +/+  g)(h  +  k) 

we  may  multiply  each  term  of  a  -\-  b  -\-  c  -{-  d  by  each  term 
of  e  +f+  ff,  then  multiply  every  product  thus  obtained  by 
each  term  of  h  +  k,  and  finally  add  the  results  of  these  last 
multiplications. 

Hence  Ave  shall  obtain  one  term  of  the  product  if  we  select 
one  term  from  each  of  the  three  given  factors  and  multiply 
these  terms  together.  And  we  shall  obtain  all  the  terms  of 
the  product  if  we  make  this  selection  of  terms  from  the  three 
given  factors  in  all  possible  ways. 

Thus,  selecting  b  from  the  first  factor,  g  f'-om  the  second, 
and  k  from  the  third,  we  have  the  term  bgk  of  the  product ; 
and  so  on. 

Since  we  can  select  a  term  from  a-\-b-{-c-\-d  in  four  ways, 
a  term  from  e  +f+  g  in  three  ways,  and  a  term  from  h  +  k 
in  two  ways,  the  number  of  terms  in  the  complete  product  is 
4  •  3  •  2,  or  24.     And  so,  in  general, 

The  product  of  any  number  of  2'>ohjnomials  is  the  sum  of  all 
the  products  that  can  be  obtained  by  selecting  one  term  from  each 
factor  and  multiplying  these  terms  together. 

And  if  the  first  factor  has  m  terms,  the  second  n,  the  third  p, 
and  so  on,  the  number  of  terms  in  the  complete  j^^oduct  —  before 
like  terms,  if  any,  have  been  collected  — is  mnp  •  •  •. 


THE   BINOMIAL   THEOREM  258 

This  theorem  supplies  a  useful  check  on  the  correctness  of  a  555 
multiplication.  It  may  be  applied  to  a  product  in  which  like 
terms  have  been  collected,  provided  its  terms  represent  sums 
of  terms  of  like  sign  and  without  numerical  coefficients,  the 
coefficient  of  a  term  then  indicating  how  many  uncollected 
terms  it  represents. 

Thus,  by  our  theorem,  the  product  {a  +  b  +  c)  {a  +  b  +  c)  should  have 
3  •  3  or  9  terms,  which  are  all  of  the  same  sign.  But,  as  we  have  shown, 
(a  +  6  +  c)  (a  +  6  +  c)  =  a2  +  62  _|.  c2  -f  2  a6  +  2  ac  +  2  6c,  and  this  repre- 
sents an  uncollected  product  of  1  +  1  +  1  +  2  +  2  +  2  or  9  terms,  as  it 
should. 

Similarly,  the  product  (a  +  b)  (a  +  b)  (a  +  b)  should  have  2  •  2  •  2  or  8 
terms.  But  this  product  when  simplified  is  a^  +  3  a-b  +  Sab-  +  b^,  which 
means,  as  it  should,  an  uncollected  product  of  1  +  3  +  3  +  1  or  8  terms. 

One  should  bear  this  theorem  in  mind  when  reckoning  with     556 
symmetric  functions  by  the  methods  of  the  last  chapter. 

Thus,  the  student  has  proved,  p.  249,  Ex.  7,  Sa6  •  2a  =  Za^b  +  3  :Eabc. 
To  test  this  formula,  suppose  that  only  the  letters  a,  b,  c  are  involved. 
Then  Za6  has  3  terms,  Sa  has  3  terms,  Sa^ft  has  6  terms,  and  2a6c  has 
1  term  ;  and  3-3  =  6  +3-1,  as  it  should. 

Products  of  binomial  factors  of  the  first  degree.     The  theorem     557 
of  §  554  enables  one  to  obtain  the  product  of  any  number  of 
factors  of  the  form  a;  +-  ^  by  inspection.     Thus, 

(x  +  h,)  (x  +  ^-,)  (x  +  b,) 

=  x^  +  (bi  +  b^  +  b,)  x""  +  (^1^.2  +  ^1^3  +  Kh)  X  +  bAK 

For,  selecting  x  from  each  factor,  we  have  the  term  x^. 

Selecting  in  all  possible  ways  x's  from  two  of  the  factors 
and  a  b  from  the  third,  we  have  the  terms  bix"^,  b^x"^,  b^x^. 

Selecting  in  all  possible  ways  x  from  one  of  the  factors  and 
b's  from  the  other  two,  we  have  the  terms  bib2X,  h^b^x,  b^b^x. 

Selecting  ft's  from  all  three  factors,  we  have  the  term  b^bj)^. 

Observe  that  when  the  terms  of  the  product  are  collected, 
as  in  the  formula,  the  coefficient  of  a-'^  is  the  sum  of  the  three 
letters  ^i,  b^,  b^,  the  coefficient  of  x  is  the  sum  of  the  products  of 


254  A   COLLEGE   ALGEBRA 

every  two  of  these  letters,  and  the  final  term  is  the  produci 
of  all  three. 

Hence  these  coefficients  are  symmetric  functions  of  h^,  b^,  b^, 
as  was  to  be  expected  since  (x  +  bi)  (x  +  b^)  (x  -f-  b^  is  itself 
symmetric  with  respect  to  b^,  b^,  b^. 

Observe  also  that  since  (x  +  b-^)  {x  +  ^g)  {^  +  ^s)  is  sym- 
metric with  respect  to  b^,  b^,  b^,  we  may  obtain  the  product 
by  finding  one  term  of  each  type,  as  x^,  b^x^,  b^b-^x,  b^b^b^,  and 
then  writing  out  all  the  terms  of  these  several  types. 

558  By  the  same  reasoning  we  may  prove  the  general  formula 

(x  +  b,)  (x  +  b.)  (x  +  b,)---  (x  +  b„) 

=  x"-^  B^x"-^  +  Box"-^  H +  5„, 

where        Bi  =  'Sbi       =  bi  +  b.^  +  b^  ^ h  b„, 

B^  =  ^b^b^    =  iA  +  bA  +  •  •  •  +  hh  +  ■■■  +  K-iK, 

Bs  =   2&1M3  =  hhh  +  bAh  +  •••+  K-2K-l^n, 

B„=  hhh---b,/, 

that  is,  Bi  is  the  sum,  and  B^  is  the  i^oduct  of  all  the  letters 
*i>  ^2>  •  •  •  ^n)  ^^^  '^^  intermediate  coefficients  are :  iJo,  the  sum 
of  the  products  of  every  txoo  of  these  letters ;  B^,  the  sum  of 
the  products  of  every  three  ;  and  so  on. 

Thus,  we  obtain  one  term  of  the  product  each  time  that 
we  select  ?/s  from  three  of  the  factors  and  o-'s  from  the  rest. 
Making  the  selection  in  all  possible  ways,  we  obtain  the  terms 
bib^bsX""^,  b^b^biX"'^,  ■  ■  -,  and  their  sum  is  Bsx""-^. 

Observe  that,  as  indicated  above,  the  coefficients 

Bi,  B^,  ■•;  B„ 

are  symmetric  fimctions  of  the  letters  bi,  bo,  •••,  h^. 

559  In  like  manner,  we  have 

(x  —  bi)  (x  -  J2)  (x  -  ^^3)  ■■■(x-  I>„) 

=  x"-  B.x"--'  +  B^x—^ +  (-  1YE„, 

where  Bi,  B^,  •■•  B„  have  the  same  meanings  as  in  §  558.  and 


THE   BINOMIAL    THEOREM  255 

the  signs  connecting  the  terms  are  alteiuately  —  and  +,  the 
last  sign,  that  of  (—  l)"i?„,  being  +  when  n  is  even  and  — 
when  n  is  odd. 

We  obtain  this  formula  by  merely  changing  the  signs  of 
all  the  letters  b^,  h^,  •••  &„  in  the  formula  of  §  558.  For  this 
leaves  unchanged  every  B  whose  terms  are  products  of  an 
even  number  of  Z»'s,  and  merely  changes  the  sign  of  every  B 
whose  terms  are  products  of  an  odd  number  of  i's. 

Example.     By  the  method  of  §§  557-559  find  the  following  products. 

1.   (X  +  1)  (X  +  2)  (a;  +  3).  2.    (x+ 2)(x  -  3)  (x  +  4). 

3.  (x4-a)(a;  +  6)(x  +  c)(x  +  d).     4.    (x  -  2/)(x  +  22/)(x  -  32/){x  +  4?/). 

The  number  of  terms  in  the  sums  2bi,  Sbibo,  •  •  •.     Let  ??i,  Wr,,  •  •  •     560 
denote  the  number  of  terms  in  2/^i,  26A>  •  •  •  respectively. 

1.  Since  2&i  =l  h-^-\-h.^ -{-•■•  A^h^,  evidently  Wi  =  n. 

2.  If  we  multiply  each  of  the  n  letters  h^,  h^,  ■•■,  i„  by  each 
of  the  other  w  —  1  letters  in  turn,  we  obtain  n  (ji  —  1)  products 
all  told.  But  these  n  (n  —  1)  products  are  the  terms  of  l.hil>2, 
each  counted  twice.  Hence  /Zg,  the  number  of  terms  in  2&i62» 
is  71  (71  —  l)/2,  or 

71—1         71  ()l  —  1) 

Thus,  we  have 

&1&25  &1&3)  •  •  -1  bibn;  b^bi,  h^z-,  •  •  •»  bibn ;  •  •  ■ ;  &«&!>  b„b2,  •  •  •,  6„6„_i. 

There  are  n  groups  of  these  products,  and  n  —  1  products  in  each 
group,  hence  n(n  — 1)  products  all  told. 

But  the  product  6162  here  occurs  twice,  namely  once  in  the  form  &162 
and  once  in  the  form  62&1 ;  and  so  on. 

3.  Again,  if  we  multiply  each  of  the  ??2  terms  of  ^b^h^  hy 
each  of  the  n  —  2  letters  which  do  not  occur  in  that  tern\ 
we  obtain  Wg  (w  —  2)  products  all  told.  But  these  ti^  (w  —  2) 
products  are  the  terms  of  ^b-fij)^,  each  cotmted  three  times. 
Hence  713,  the  number  of  terms  in  ^b^h.jb^,  is  n2{7i  —  2)/3,  or 

71-2      n  (71-1)  {71 -2) 


256  A   COLLEGE   ALGEBRA 

Thus,  we  have 
6i&2&3)  bibibi,  •  ■  •,  bib2bn;  &1&3&2)  bib^bi,  •  •  •,  ^i^s^nj 

•  •  • ;  b„-ib„bu  bn-ib„b2,  ■  ■•,  6a_i6A-2. 

There  are  n^  groups  of  these  products,  and  n  —  2  products  in  each 
group,  hence  n^  (n  —  2)  products  all  told. 

But  the  product  616263  here  occurs  three  times,  namely  in  three  forms, 
6162  •  63,  6163  •  62,  6263  •  61.  And  similarly  every  term  of  S616263  here 
occurs  three  times,  once  for  each  of  the  three  ways  in  which  a  product 
of  three  letters  may  be  obtained  by  multiplying  the  product  of  two  of 
the  letters  by  the  remaining  letter. 

4.    By  the  same  reasoning  we  can  show  that 

n-3  _n(n-l)(n-  2)  (n  -  3) 
'''-'''     4      ~  1.2-3.4 

and,  in  general,  that 

n(n  —  l)(n  —  2)  ■••  to  r  factors 
"'-'- 1.2.3...,- 

Thus,  the  numbers  of  products  of  four  letters  6i,  62,  63,  64,  taken  ov,e, 
two,  three,  four  at  a  lime,  are 

4-3      ^  43 -2      ,  4-3.2-1       . 

„i  =  4,     n2  =  — =  6,     n3  =  j^:^  =  4,     n,  =  ^-^-^  =  L 

561         Binomial  theorem.     If  in  the  formula  of  §  558,  namely, 

(x  +  bi)  (x  +  b^)---  (x  +  b„)  =  ic"  +  Bix"-^  +  B^x''--  -\ \- B„, 

we  replace  all  the  n  different  letters  bi,  b^,  ■  ■  ■  b„  by  the  same 
letter  b,  and  x  by  a,  the  first  member  becomes  (a  +  b)". 

Again,  since  each  of  the  71  terms  of  B^  becomes  b,  and  each 
of  the  Tiz  terms  of  B^  becomes  b"^,  and  so  on,  we  have,  §  560, 

n   -nb    B   -^^("-1)/,.     n      _n(n-l)(n-2-) 

■Di  —  no,  n^—       J    ,^       0,  li^—  103  <>,•••. 

Our  formula  therefore  reduces  to  the  following : 

(a  +  &)»  =  a"  +  J  a^-'b  +  '^^^^  a-^' 

+  1-2.3  «      6  +   •. 

where 


THE   BINOMIAL    THEOREM  257 

1.  The,  7iumber  of  terms  on  the  right  is  n  -)-  !• 

2.  The  exponents  of  a,  decrease  by  one  and  those  ofh  increase 
by  one  from  term  to  term,  their  sum  in  each  term  being  n. 

3.  The  first  coefficient  is  1,  the  second  is  n,  and  the  rest  of 
them  may  be  found  by  the  following  rule: 

Multiply  the  coefficient  of  any  term  by  the  exponent  of  a  in 
the  term  and  divide  by  the  exponent  of  b  increased  by  1 ;  the 
result  will  be  the  coefficient  of  the  next  term. 

This  formula  is  known  as  the  binomial  theorem,  the  expres- 
sion on  the  right  being  called  the  expansion  of  {a  +  ^)"  by  this 
theorem. 

•Thus,  (a  +  6)3  =  a3  +  3a26  +  — a62  +  ?_L?ji53 

^  '  1-2  1.2-3 

=  a3  +  3  a?-h  +  3  a62  +  W. 

Since  a  +  ^  is  symmetric  with  respect  to  a  and  b,  it  follows  562 
from  §  542  that  the  terms  of  the  same  type  in  the  expansion  of 
{a  +  Z»)"  —  namely  those  involving  a."  and  i",  a"~'6  and  ab"~^, 
and  so  on  —  must  have  the  same  coefficients.  But  these  are 
the  first  and  last  terms,  the  second  and  next  to  last  terms,  and 
in  general  every  two  terms  which  are  equally  removed  from 
the  beginning  and  the  end  of  the  expansion. 

Hence  the  last  term  is  b",  the  next  to  last  is  nab"~^,  and  so 
on.  Since  the  number  of  terms  is  n  +  1,  there  will  be  one 
middle  term  when  n  is  even,  two  when  n  is  odd.  When  there 
are  two  middle  terms,  they  are  of  the  same  type  and  have  the 
same  coefficients.  And  by  what  has  just  been  said,  the  coeffi- 
cients of  the  terms  which  follow  the  middle  term  or  terms  are 
the  same  as  those  which  precede  them but  in  reverse  order. 

It  may  also  readily  be  shown  that  the  coefficients  increase     563 
up  to  the  middle  term  or  terms  and  then  decrease,  so  that 
the  middle  coefficient  or  coefficients  are  the  greatest. 

This  follows  from  the  rule  of  coefficients,  §  561,  3,  since  in  the  terms 
which  precede  the  middle  term  or  terras  the  exponent  of  a  is  greater 
than  the  exponent  of  b  increased  by  1,  while  in  the  terms  which  follow 
it  is  less. 


258  A   COLLEGE   ALGEBRA 

564  By  changing  the  sign  of  h  in  the  preceding  formula  and 
simplifying,  we  obtain 

(a  -  hy  =  a"  -  na^-'b  +  ^^^^^  a-'b' 

n(n-l)(n-2) 

1-2.3  «      ^  +      » 

where  the  terms  which  involve  odd  powers  of  b  have  —  signs^ 
and  those  which  involve  even  powers  have  +  signs. 

Example.     Find  the  expansion  of  (2  x  -  y^)^. 

Substituting  2  x  for  a,  and  y^  for  6,  in  the  formula,  and  remembering 
that  the  last  three  coefficients  are  the  same  as  the  first  three  in  reverse 
order,  §  562,  we  have 
(2  X  -  2/3)6  ^  (2  x)6  -  6  (2  x)^y^  +  ^  (2x)*  (t/3)2  _  ^lllA  (2  x)^  (2/3)3  + . . . 

=  (2  x)6  -  6  (2  x)52/3  +  15  (2  x)*  {yY  -  20  (2  x)3  {y^ 

+  15  (2  X)'^  (7/3)4  _  6  (2  X)  (2/3)5  +  (y3)6 

=  64x6  -  192x^2/3  +  240x*2/e  -  160x3j/9  +  mx'^y^'^  -  12x2/^5  +  y". 

565  The  general  term.     From  §  561  it  follows  that  the  (r  -f  l)th 
term  in  the  expansion  of  {a  +  by  is 

n(n  —  V)  (n  —  2)  ■  •  ■  to  r  factors 

■ — ^ -^ ' a^~^b^. 

1  •  2  •  3  •  •  •  r 

This,  with  a  minus  sign  before  it  when  r  is  odd,  is  also  the 
(r  +  l)th  term  in  the  expansion  of  {a  —  by. 

Example  1.     Find  the  eighth  term  in  the  expansion  of  (x  —  yy^. 
Here  n  =  16  and  r  +  1  =  8,  or  r  =  7.     Hence  the  required  term  is 
16 -15 -14    13    12    11    10 


1.2.3-4-5-6-7 


x^z/T  =  -  11440x92/'- 


Example  2.     Does  any  term  in  the  expansion  of  (x3  +  l/x)i2  contain 
x2f»  ?     If  so,  find  this  term. 

Let  r  +  1  denote  the  number  of  the  term.    Then,  since  n  =  12,  a  =  x*, 
and  b  =  1/x,  vre  mu.st  have 

a'>-'-b'-  =  (x3)i2-'-/x'-  =  x^-*'-  =  x2o. 


THE   BINOMIAL   THEOREM  259 

This  condition  is  satisfied  if  36  —  4  r  =  20,  or  r  =  4. 

Hence  the  fifth  term  contains  x^o,  and  substituting  in  the  formula  we 

find  this  term  to  be  lUijJ^a;2o  =  495  x2o. 
1-2. 3-4 


EXERCISE  XXXI 

Expand  the  following  by  aid  of  the  binomial  theorem. 
1.   (3x  +  2  2/)3.  2.   (a -6)8.  3.   (1  +  2x2)7. 

4.   (2  +  l/x)4.  5.   {x-3/x)6.  6.   (x/y-y/z)^ 

7.    (1  -  X  +  2  x2)*.  8.   (a2  +  ax  -  x2)3. 

9.    Find  the  sixth  term  in  (1  +  x/2)". 

10.  Find  the  eighth  term  in  (3  a  -  4  6)i2, 

11.  Find  the  middle  term  in  (a^  -  2  bcy°. 

12.  Find  the  two  middle  terms  in  (1  —  x)^. 

13.  Find  the  coeflicient  of  x^  in  (1  +  x)^. 

14.  Find  the  coefficient  of  x*  in  (3  -  2  xy. 

15.  Find  the  coefficient  of  x^  in  (1  —  x-)^. 

16.  Find  the  coefficient  of  x^  in  (1  +  2  x)^  +  (1  -  2«)". 

17.  Find  the  constant  term  in  (x  +  l/x)i2. 

18.  Find  the  coefficient  of  x"^  in  (2  x  —  1  /x)  is. 

19.  Find  {x  +  2y)  {x  —  Sy)  {x  —  5y)  by  inspection. 

20.  Find  (x  +  2)  (x  +  3)  (x  -  4)  (x  -  5)  by  inspection. 

21.  What  is  the  number  of  terms  in  the  product 

{a  +  b  +  c  +  d){f+g  +  h){k  +  l)  {m  +  n+p  +  q)? 

22.  Find  the  sum  of  the  coefficients  in  the  following  products. 

1.     (i  +  X2  +  X3  +  X4)3.  2.     (1  +  2  X  +  X2)2  (1  +  X  +  3  X^)^. 

23.  What  is  the  sum  of  the  coefficients  in  the  following  symmetric 
functions  of  four  letters  a.  6,  c,  d  when  expanded  ? 

1.    Sa2  •  2a.  2.    2,a^  ■  "Slabc.  3.    Sa6  •  Zabc. 

24.  Show  that  the  sum  of  the  coefficients  in  the  expansion  of  (a + 6)"  is  2". 

25.  Show  that  in  the  expansion  of  (a  —  6)"  the  sum  of  thg  positive 
coefficients  is  numerically  equal  to  the  sum  of  the  negative  coefficients. 


260  A   COLLEGE   ALGEBRA 


XL     EVOLUTION 


566  Perfect  powers.  Given  a  rational  function  P.  It  is  possible 
that  P  is  a  lyerfeot  nth  power;  in  other  words,  that  a  second 
rational  function  Q  exists  such  that  P  —  Q".  If  so,  this  rational 
function  Q  will  be  an  nth  root  of  P. 

In  the  present  chapter  we  consider  the  problem  :  A  rational 
function  P  being  given,  it  is  required  to  determine  whether  or 
not  P  is  a  perfect  nth  power,  and,  if  it  is,  to  find  its  7ith  root  Q. 
We  suppose  n  to  denote  a  given  positive  integer. 

567  Roots  of  monomials.  Let  P  denpte  a  rational  monomial 
reduced  to  its  simplest  form.  If  P  is  a  perfect  nth  power, 
an  7ith  root  of  P  may  be  obtained  by  the  following  rule. 

Divide  the  exj)onents  of  the  several  literal  factors  of  P  by  n, 
and  multiply  the  result  by  the  principal  nth  root  of  the  numeri- 
cal coefficient  ofP. 

This  follows  at  once  from  the  rule  for  involution,  §  318. 
Thus,   (aW/c'«)"  =  ai''6'''/t"'",   §318.     Hence  a'^b'/C"  is  an  nth  root 
of  a'-"V"/c""\   §  566,  and  it  is  obtained  by  dividing  the   exponents  of 

568  The  root  thus  obtained  is  called  the  principal  nth  root  of  P 
(compare  §  258).  We  shall  mean  this  root  wh^n  we  speak  of 
the  nth  root  of  P,  or  when  we  use  the  symbol  VP. 

Example  1.     Find  the  cube  root  of  -  8  a^h^/21  x^y^. 


We  have  a/ 


-  8  a%^  _      2  ab^ 


Example  2.     Find  the  following  roots, 
1 


04  a*66 


^^Ixhfz^K  3. 


6/32^ 


100  c8di2 
569         Roots  of  polynomials.     Consider  the  following  examples. 

Exainp^le  1.     Determine  wliether  or  not  4  a;*  —  4  a;^  +  13  x2  —  6  a;  + 
a  perfect  square,  and,  if  it  is,  find  its  square  root. 


EVOLUTION  261 

If  this  is  a  perfect  square,  evidently  it  must  have  a  square  root  of  the 
form  2  x'^  +px  +  q,  where  p  and  q  are  constants.     We  must  therefore  have 

Ax*  -  4x^  +  rSx^  -  6x  +  9  =  (2x'^  +  px  +  qf 

=  4x*  +  4jjx3  +  (p-  +  ■iq)x'^  +  2pqx  +  q'^, 
which  requires,  §  284,  that  p  and  q  satisfy  the  equations : 
4p  =  -4     (1),       2)2  +  4g  =  13     (2),        2pg  =  -6     (3),        q^  =  9     (4). 

From  (1)  and  (2)  we  find  p  =  -  1,  q  =  3,  and  these  values  of  p  and  q 
satisfy  (3)  and  (4) ;  for  2  (-  1)  3  =  -  6,  and  3^  =  9. 

Hence  4  x'*  -  4  x-^  +  13  x-  -  6  x  +  9  is  a  perfect  square  and  2  x^  -  x  +  3 
is  its  square  root. 

Example  2.     Find  the  cube  root  of 

x6  +  6x5  +  21x*  +  44 x^  +  63x2  f  54x  +  27. 
If  this  be  a  perfect  cube,  it  will  have  a  cube  root  of  the  form  x^+px  +  q. 
We  must  therefore  have 

x6  +  6  x5  +  2 1  X*  +  44  x3  +  63  x2  +  54  X  +  27  =  (x2  +  px  +  qf 
=  x6  +  3px6  +  3(p2  +  q)xi  +  (p3  +  epq)x^ 
+  3  {p-q  +  g2)  x2  +  3  J992^  +  q^, 
which  requires,  §  284,  that  p  and  q  satisfy  the  six  equations : 

3p  =  6     (1),  3(p2  +  g)  =  21     (2),-..  gs  ^  27     (6). 

From  (1)  and  (2)  we  obtain  p  =  2,  g  =  3.  And  these  values  of  p  and  q 
will  be  found  to  satisfy  the  remaining  equations  (3)-  •  -(6). 

Hence  x^  +  6x5  +  •  •  •  +  54x  +  27  is  a  perfect  cube,  and  its  cube  root 
is  x2  +  2  X  +  3. 

By  the  method  illustrated  in  these  examples  it  is  always 
possible  to  determine  whether  or  not  a  given  polynomial  in  x 
is  a  perfect  nth.  power,  and,  if  it  is,  to  find  its  nth.  root. 

Let  the  polynomial  be  aox"'  +  aja;"'"^  +  •  •  •  +  «„,.  If  this  be 
a  perfect  wth  power,  its  degree  m  must  be  a  multiple  of  n  so 
that  m  =  kn,  where  k  is  an  integer ;  and  it  must  have  an  nth 
root  of  the  form  ax''  +  A^x'-'^  -\-  ■  ■  •  +  A^.,  where  a  denotes  the 
principal  wth  root  of  ao,  and  -4i,  •  •  ■,  Ai.  are  unknown  constants. 
We  call  this  root  the  principal  nt^i  root. 

To  determine  whether  OoX"'  +  •  ■  •  +  a,„  has  any  such  root, 
and  to  find  this  root  if  it  exists,  we  set 

aoX'"  +  a^x'"-'  4-  . . .  +  a„  =  (ax'  +  A,x'-'  +  ■■■  +  A,)". 


262  A   COLLEGE   ALGEBRA 

/  reduce  the  second  member  to  the  form  of  a  polynomial  in  x, 
and  then  equate  its  coefficients  to  those  of  the  like  powers  of 
X  in  the  first  member.  We  thus  obtain  a  system  of  nk  equa- 
tions in  A^,  Ao,  ■■■,  A^.  The  first  k  of  these  equations  will 
give  a  single  set  of  values  for  A^,  An,  ■■■,  A^\  and  this  set  of 
values  must  satisfy  the  rest  of  the  equations  if  a^x™  +  •  ■  ■  -\-  a^ 
is  to  be  a  perfect  ?ith  power. 

Example  3.     Find  the  cube  root  of 

8  x6  -  12  x5  +  18  x«  -  13  x3  -f  9  x2  -  3  X  +  1. 

570  Square  roots  of  polynomials.  If  a  polynomial  P  is  a  perfect 
square,  its  square  root  may  also  be  obtained  by  the  following 
method. 

As  in  the  preceding  section,  let  P  denote  a  polynomial  in  x 
of  even  degree  and  arranged  in  descending  powers  of  x. 

Let  us  suppose  that  P  is  a  perfect  square  and  that  a,  b,c,  ■•• 
denote  the  terms  of  its  square  root  arranged  in  descending 
powers  of  x,  so  that  P  =  (a+J  +  c  +  --  -y. 

The  problem  is,  knowing  P,  to  find  a,  b,  c,  ■■  ■. 

Now,  whatever  the  values  of  a,  b,  c,  ■■•  may  be,  we  have 

(a  +  by  =  a^  +  2ab  +  b^  =  a^  +  (2a  +  b)b, 
{a  +  b  +  cY={a  +  by  +  2  {a  +  b)c  +  c"" 

=  a'+{2a  +  b)b+[_2{a  +  b)  +  c-]c, 
{a  +  b  -{-  c  -\-  dy  ==  a^  +  {2 a  +  b)b  +[2{a  +  b)  +  c-]c 
+  [2{a  +  b  +  c)+cqd, 

and  so  on,  a  new  group  of  terms  being  added  on  the  right  with 
each  new  letter  on  the  left,  namely,  a  group  formed  by  addimj 
the  tieio  letter  to  twice  the  sum  of  the  old  letters  and  multijjlying 
the  result  by  the  new  letter. 

Therefore,  since  by  hypothesis  P  =  [a  -{-  b  +  c  +  •  •  -y,  we 
have 

P  =  a'+(2a  +  b)b+{_2{a  +  b)+  (f\c 

+  [2(a  +  6  +  c)  +  o?](^  +  ..., 


EVOLUTION  263 

where  the  leading  terms  of  the  several  groups  on  the  right, 
namely,  d\  2  ah,  2,  ac,  2ad,---,  are  all  of  higher  degree  in  x 
than  any  of  the  terms  which  follow  them. 

From  this  identity  we  may  find  a,  b,  c,  ■  ■  ■  as  follows  : 

1.  Evidently  a  is  the  square  root  of  the  leading  term  of  P. 

2.  Subtract  a^  from  P.  As  the  leading  term  of  the  remainder, 
Ri,  must  equal  2  ab,  we  may  find  b  by  dividing  this  term  by  2  a. 

3.  Having  found  b,  form  {2a  +  b)b  and  subtract  it  from  Ri. 
As  the  leading  term  of  the  remainder,  R^,  must  equal  2  ac,  we 
may  find  c  by  dividing  this  term  by  2  a. 

4.  Continue  thus  until  a  remainder  of  lower  degree  than  a 
is  reached. 

If  this  final  remainder  is  0,  P  is,  as  was  supposed,  a  perfect 
square  and  its  square  root  is  a  +  Z>  +  c  +  •  ■  •. 

If  this  final  remainder  is  not  0,  P  is  not  a  perfect  square ; 
but  we  shall  have  reduced  P  to  the  form 

P=(a  +  b  +  c  +  .--y+R, 
that  is,  to  the  sum  of  a  perfect  square  and  an  integral  function 
which  is  of  lower  degree  than  a. 

It  is  convenient  to  arrange  the  reckoning  just  described  as  in 
the  following  example. 

Example.     Find  the  square  root  of  4  x*  -  4  z^  +  13  a;2  _  g  ^  +  9. 

P  =  4x*-4x3  +  13x2-6x  +  9  |2x'^-x  +  3  =  a  +  6  +  c 


a^  =  4x* 

2a  +  6  =  4x2-x 

-  4x3  +  13x2  -  6x  +  9  =  El  =  P  -  a2 

-  4 x3  +      x2                   =i2a  +  b)b 

2(a  +  6)  +  c  =  4x2- 

-2x  +  3 

12x2  -  6x  +  9  =  E2  =  P  -  (a  +  6)2 
12  x2  -  6  X  +  9  =  \2{a  +  b)  +  clc 

0              =  E  =  P  -  (a  +  6  +  c)2 

Since  the  final  remainder  is  0,  P  is  a  perfect  square  and  its  square 
root  is  2x2  -  X  +  3.     Compare  §  569,  Ex.  1. 

Observe  that  as  each  new  remainder  R^,  R^,  ■••  is  found  we 
divide  its  leading  term  by  2  a  and  so  get  the  next  term  of  the 
root.  Then  at  the  left  of  the  remainder  we  write  twice  the 
part  of  the  root  previously  obtained  plus  the  new  term  of 


264  A   COLLEGE   ALGEBRA 

the  root.  We  multiply  this  smn  by  the  new  term  of  the  root, 
subtract  the  result  from  the  remainder  under  consideration, 
and  thus  obtain  the  next  remainder. 

Example.     Find  the  square  root  of  25  x*  —  40x3  +  46  x'^  -  24x  +  9. 

571  This  method  is  applicable  to  a  polynomial  P  which  involves 
more  than  one  letter,  provided  it  be  a  perfect  square.  We 
first  arrange  P  in  descending  powers  of  one  of  the  letters, 
with  coefficients  involving  the  rest,  and  then  proceed  as  in 
§  570,  it  being  understood  that  x  now  denotes  the  letter  of 
arrangement. 

572  Approximate  square  roots.  We  may  also  apply  this  method 
to  a  polynomial  in  x  arranged  in  ascending  powers  of  this 
letter.  But  a,  b,  c,  ■  ■■  will  then  be  arranged  in  ascending 
powers  of  x,  and  the  degrees  of  the  successive  remainders  will 
increase.  Hence,  §  570,  4,  if  P  is  not  a  perfect  square  but  has 
a  constant  term,  we  can  reduce  it  to  the  form 

that  is,  to  the  sum  of  a  perfect  square  and  a  polynomial,  R', 
whose  lowest  term  is  of  as  high  a  degree  as  we  jjlease. 

For  small  values  of  x  we  can  make  the  value  of  it'  as  small 
as  we  choose  by  carrying  this  reckoning  far  enough.  Hence 
in  this  case  we  call  a  +  i,  a  +  ^  +  c,  •  •  •  the  approximate  square 
roots  of  P  to  two  terms,  three  terms,  and  so  on. 

It  should  be  added  that  these  approximate  roots  are  found 
more  readily  by  the  method  of  §  569. 

Example  1.     Find  the  square  root  of  1  +  x  to  four  terms. 
By  §  569,  we  write  Vl  +  x  =  1  +  px  +  ^x^  +  rx^  +  •  •  ■ . 

Squaring,  1  +  x  =  1  +  2px  +  (p^  +  2  q)  x"-  +  2{j>q  +  r)  x^  +  .  •  • 

Hence,  §284,      2p=l,    ^2  +  2^  =  0,    pg  +  r  =  0, 
or  solving,  p  =  l/2,  q'=— 1/8,     r=l/16. 

Therefore  the  required  result  is  1  +  x/2  —  x'-/8  +  x^/lO. 
Let  the  student  verify  this  by  the  method  of  §§  570,  57  L 

Example  2.     Find  the  square  root  of  4  —  x  +  x^  to  three  terms. 


EVOLUTION  265 

Square  roots  of  numbers.     From  the  formulas  in  §  570  we     573 
also  derive  the  ordinary  method   of  finding  the  square  root 
of  a  number. 

Example.     Find  the  square  root  of  533(31. 

Let  a  denote  the  greatest  integer  with  but  one  significant  figure,  whose 
square  is  contained  in  5336 L  Its  significant  figure  will  be  the  leading 
figure  of  the  root  and  its  remaining  figures  wiil  be  O's.  We  find  a  as 
follows : 

Remembering  that  for  each  0  at  the  end  of  a  there  will  be  two  O's  at 
the  end  of  a-,  we  mark  off  in  53361,  from  right  to  left,  as  many  periods 
of  two  figures  as  we  can,  thus  :  5'33'61. 

Each  of  the  periods  61  and  33  calls  for  one  0  at  the  end  of  a,  and  the 
remaining  period,  5,  calls  for  the  initial  figure  2,  since  2  is  the  greatest 
integer  whose  square  is  less  than  5.     Hence  a  —  200. 

Having  found  a,  we  proceed  quite  as  when  seeking  the  square  root  of 
a  polynomial.  This  is  indicated  in  the  scheme  below  at  the  left,  where  b 
denotes  the  second  figure  of  the  root  multiplied  by  10,  and  c  the  units 
figure.  The  scheme  at  the  right  gives  the  reckoning  as  abridged  in 
common  practice. 

a   +  b  +  c 
'5' 33' 61 1 200 +  30+1 
4  00  00  =  a^ 
2a  =  400 


2a  +  6  =  430 


1  33  61  =  Ri 

1  29  00  =  (2  a  +  6)  6 


2  (a +  6)  =  460  4  61  =  R2 
2 (a  +  ?>)  +  c  =  461 14  61  =  [2(a  +  b)  +  c]c 
0     =  R 

We  first  subtract  a^,  then  find  the  significant  figure  of  b  by  dividing 
the  remainder  i?i  by  2  a,  next  find  Ro  by  subtracting  {2a  +  b)b  from  Ri, 
and  finally  c  by  dividing  R2  by  2  (a  +  b). 

The  simplest  way  of  accomplishing  all  this,  as  indicated  in  the  abridged 
scheme  at  the  right,  is  to  omit  final  O's  and  to  bring  down  one  period 
at  a  time.  Then,  as  each  new  remainder  is  obtained,  we  write  at  its 
left  twice  the  part  of  the  root  already  found  as  a  "trial  divisor,"  obtain 
the  next  figure  of  the  root  by  dividing  the  remainder  by  this  trial  divisor, 
and  complete  the  divisor  by  affixing  this  figure  to  it.  We  then  multiply 
the  complete  divisor  by  the  new  figure  of  the  root,  subtract,  and  so 
obtain  the  next  remainder.  If  too  large  a  figure  is  obtained  at  any  stage 
•n  the  process,  that  is,  a  figure  which  makes  the  product  just  described 
greater  than  the  remainder  in  question,  we  try  the  next  smaller  figure. 


266  A   COLLEGE   ALGEBRA 

574  Approximate  square  roots  of  numbers.  The  method  just 
explained  also  enables  us  to  obtain  approxhnate  values  of 
the  square  roots  of  numbers  which  are  not  perfect  squares. 

Example.  Find  an  approximate  vahie  of  the  square  root  of  7.342  cor- 
rect to  the  third  place  of  decimals. 

7.34'20'00I2.709  Evidently  for  each  decimal  fig- 

' ure  m  the  root  there  are  two  such 

.  _  -  o  figures  in  the  number.     Hence  we 

„  90  Hence  separate  the  decimal  part  of  the 

54^^  20  00       V7:M  =  2.709...     ™^''  into  periods   of   two  fig- 

.  ures,  proceednig  from  the  decimal 

1 — ——  point  to  the  right.     The  integral 

part  of  the  number  we  separate  into 

periods,  as  in  §  573,  proceeding  from  the  decimal  point  to  the  left. 

Observe  that  a  decimal  number  cannot  be  a  perfect  square 
if  it  has  an  odd  number  of  decimal  figures. 

575  Cube  roots  of  polynomials.  There  is  also  a  special  method  for 
finding  the  cube  root  of  a  polynomial  P,  when  P  is  a  perfect 
cube,  analogous  to  that  just  given  for  finding  a  square  root. 

Let  P  denote  a  polynomial  in  x  whose  degree  is  some  multi- 
ple of  3  and  which  is  arranged  in  descending  powers  of  x. 

Let  us  suppose  that  Pisa  perfect  cube  and  that  a,b,c,-'- 
denote  the  terms  of  its  cube  root,  arranged  in  descending 
powers  of  x,  so  that  P  =  (a-\-h-\-c-{---  -y. 

The  problem  is,  knowing  P,  to  find  a,  b,  c,  ■■  -. 

Now,  whatever  the  values  of  a,  l>,  c,  ■■■  may  be,  we  have 
(a  +  by=  a^  +  (3  fl2  +  3  ai  +  b"")  b, 
(a  +  i  +  c)3  =  a"  +  (3  «2  +  3ab  +  b^)b 

+  [3(a  +  by  +  3(«  +  b)c  +  c2]c, 
and  so  on,  a  new  group  of  terms  being  added  on  the  right  with 
each  new  letter  on  the  left,  namely,  a  group  formed  bij  adding 
together  three  times  the  square  of  the  sum  of  the  old  letters,  tJiree 
times  the  product  of  the  sum  of  the  old  letters  by  the  new  letter, 
and  the  square  of  the  new  letter,  and  then  multiplying  the  result 
by  the  new  letter. 


EVOLUTION  267 

Therefore,  since  by  hypothesis  P  =  (a  +  6  +  c  +  --  •)^,  we 
have 

P  =  a^  +  (Sa""  +  3  ab  +  b^)b  +  [3  (a  +  by  +  3  (a  +  b)c  +  c^^c+  •  ■  ; 
where  the  leading  terms  of  the  several  groups  on  the  right, 
namely,  a^,  3  a^b,  3  a'^c,  •  •  •,  are  all  of  higher  degree  in  x  than 
any  of  the  terms  which  follow  them. 

From  this  identity  we  may  find  a,  b,  c,  •••  as  follows  : 

1.  Evidently  a  is  the  cube  root  of  the  leading  term  of  P. 

2.  Subtract  a^  from  P.  As  the  leading  term  of  the  remain- 
der, Ri,  must  equal  3  a^,  we  may  find  b  by  dividing  this  term 
by  3  a^. 

3.  Having  found  b,  form  (3  a^  -\-  3  ab  +  b^)  b  and  subtract  it 
from  Ri.  As  the  leading  term  of  the  remainder,  R2,  must 
equal  3  a^c,  we  may  find  c  by  dividing  this  term  by  3  a\ 

4.  Continue  thus  until  a  remainder  is  reached  which  is  of 
lower  degree  than  a"^. 

If  this  final  remainder  is  0,  then  P  is,  as  was  supposed,  a 
perfect  cube  and  its  cube  root  is  a  +  ^  +  c  +  •  ■  • . 

If  this  final  remainder  is  not  0,  P  is  not  a  perfect  cube,  but 
we  shall  have  reduced  it  to  the  form 

p  =  (a  +  b  +  c-] y  -h  R, 

where  R  is  of  lower  degree  than  a^. 

It  is  convenient  to  arrange  this  reckoning  as  follows : 

Example.     Find  the  cube  root  of 

x6  +  6x6  +  21x*  +  44x3  +  63x2  ^  54 3;  +  27. 

|x2  +  2x  +  3 
3a2  =  3x*     x6  + 6x5  +  21x^  +  44x3  +  63x2  +  ,54x  +  27 


3(x2)2  =  3x* 
3x2-2x  +  (2x)2z=6x3  +  4x2 


Ix*  +  6x3  +  4x2 


6  x5  +  21  X*  +  44  x3  +  63  x2  +  64  X  +  27 
6x5  +  12x4+    8x3 


3{x2+  2x)2  =  3x*  +  12x3+  12x2 
3(x2  +2x)3  +  32=       9x2  +  18x  +  9 


;x*  +  12x3  +  21x2  +  18x  + 


f)x*  +  36x3  +  63x2  +  54x  +  27  =  Rj 

Qx"  +  36x3  +  63x2  +  54x  +  27 

0  =R 


268  A    COLLEGE    ALGEBRA 

Since  the  final  remainder  is  0,  x^  +  6  x^  +  •  •  •  +  54  x  +  27  is  a  perfect 
cube  and  its  cube  root  is  x-  +  2  x  +  3.     Compare  §  569,  Ex.  2. 

Observe  that  as  each  new  remainder  R^,  Ri,  •  ••  is  found,  we 
divide  its  leading  term  by  3  a^  and  so  get  the  next  term  of 
the  root.  Then  at  the  left  of  the  remainder  we  write  the  sum 
of  three  times  the  square  of  the  part  of  the  root  previously 
obtained,  three  times  the  product  of  this  part  by  the  new 
term,  and  the  square  of  the  new  term.  We  multiply  this 
sum  by  the  new  term,  subtract  the  result  from  the  remainder 
under  consideration,  and  thus  obtain  the  next  remainder. 

576  This  method  is  also  applicable  to  a  polynomial  which  involves 
more  than  one  letter,  if  it  be  a  perfect  cube  (compare  §  571). 

The  method  may  also  be  applied  to  a  polynomial  in  x 
arranged  in  ascending  powers  of  this  letter,  —  if  it  does  not 
lack  a  constant  term.  If  the  polynomial  is  not  a  perfect  cube, 
we  thus  obtain  approximate  cube  roots  (compare  §  572). 

577  Cube  roots  of  numbers.     We  may  also  find  the  cube  root  of 
a  number  by  aid  of  the  formulas  of  §  575. 

Example.     Extract  the  cube  root  of  12487168. 

a    +   6  +  c 
N  =  12'  487'  168  [200  +  30  +  2  =  232 
8  000  OOP 
3a2  =  120000 


Zah=    18000 
62  =       900 


138900 


4  487  108  =  Ri  =  2f 


4  167  OOP  =  (3  a2  +  3  a6  +  b^-)  b 


1  (a  +  6)2  =  158700 
l(a  +  6)c=      1380 

c2  = 4 

160084 


320  168  =  E2  =  iV  -  (a  +  bf 


320  168  =  [3  (a  +  6)2  +  3  (a  +  6)  c  +  c"-]  c 


0        =R  =  N-{a  +  b  +  c)\ 


In  order  to  find  a,  the  greatest  number  with  one  significant  figure 
whose  cube  is  contained  in  N,  we  begin  by  marking  off  periods  of  three 
figures  in  N  from  right  to  left  (also  from  the  decimal  point  to  the  right 
when  there  are  decimal  figures  in  N),  thus  :  12'  487'  168.  Each  of  the 
periods  168  and  487  calls  for  one  0  at  the  end  of  a,  and  the  remaining 


EVOLUTION  269 

period,  12,  calls  for  the  initial  figure  2,  2  being  the  greatest  integer  whose 
cube  is  contained  in  12.     Hence  a  =  200. 

The  rest  of  the  reckoning  is  fully  indicated  above. 

Observe  that  each  new  figure  of  the  root  is  found  by  dividing  the 
remainder  last  obtained  by  three  times  the  square  of  the  part  of  the  root 
already  found ;  thus,  we  find  the  significant  figure  of  b  by  dividing  jRi  by 
3  a-,  and  c  by  dividing  2?2  by  3  (a  +  b)'^.  If  too  large  a  figure  is  thus 
obtained,  we  test  the  next  smaller  figure. 

The  process  may  be  abbreviated  in  the  same  way  as  that  for  finding 
the  square  root  of  a  number. 

Approximate  cube  roots  of  numbers  which  are  not  perfect 
cubes  may  also  be  found  by  this  process  (compare  §  574). 

Higher  roots  of  polynomials.     The  fourth  root  of  a  polynomial     578 
which  is  a  perfect  fourth  power  may  be  obtained  by  finding 
the  square  root  of  its  square  root ;  similarly  the  sixth  root  of 
a  polynomial  which  is  a  perfect  sixth  power  may  be  obtained 
by  finding  the  cube  root  of  its  square  root. 

It  is  also  possible  to  develop  special  methods,  analogous  to 
those  of  §§  570,  575,  for  finding  any  root  that  may  be  required. 
But  the  general  method  of  §  569  makes  this  \mnecessary.  In 
fact  we  have  given  the  special  methods  for  square  and  cube 
roots  explained  in  §§  570,  575  only  because  of  their  historic 
interest  and  their  relation  to  the  problem  of  finding  square 
and  cube  roots  of  numbers. 

EXERCISE   XXXn 

Simplify  the  following  expressions. 


3/      27xe^_  /5i9^_  ^^_^^_^^ 

\      125a92i2  \625c2d8  ^    ^  y   ^     y/ 

By  §  569  or  §  570  find  the  square  roots  of  the  following. 

4.  X*  -2x3  +  3x2  -2x+  1. 

5.  x2  -  2  X*  +  6  x3  -  6  X  +  x6  +  9. 

6.  4 x6  +  12 x^y  +  9x4?/2  -  4  x^yS  _  6 xV  +  2/®- 

7.  4»2_20x  +  13  +  30/x  +  9/a;2. 


270  A   COLLEGE   ALGEBRA 

8.  49 -84a; -34x2  + 60x8 +  25x*. 

9.  x8  +  2  x^  -  x6  -  X*  -  6  x3  +  5  x2  -  4  X  +  4, 

10.  (X2  +  1)2_4X(X2-1). 

11.  4x*  +  9x2y2  _  12x32/  +  16x2  -  24x2/  +  16. 

12.  X2/2/2  +  2/2/X2  +  2  +  2  X2  +  2  y2  +  x22/2. 

Find  approximate  square  roots  to  four  terms  of  the  following. 

13.  l-2x.  14.   4-x  +  3x2. 
By  §  569  or  §  575  find  the  cube  roots  of  the  following. 

15.  x6  +  3  x5  +  6  x*  +  7  x3  +  6  x2  +  3  X  +  1. 

16.  27  xi2  +  27  xio  -  18  x8  -  17  x6  +  6  X*  +  3  x2  -  1. 

17.  8  x6  -  36  ax5  +  90  aH*  -  135  aH^  +  135  a*x2  -  81  a^x  +  27  a^. 

18.  x^/y^  +  2/Va;3  +  Zx'-/y^  +  3y2/a;2  +  Q^/y  +  62//X  +  7. 

19.  Find  the  approximate  cube  root  to  three  terms  of  the  expression 

1  -  X  +  X2. 

20.  By  §  569  or  §  578  find  the  fourth  root  of 

x8  -4x^  +  10x6-  16x5+  19x*-  16x3  + 10x2 -4x  +  1. 

21.  By  §  569  find  the  fifth  root  of 

a;io  +  5a;9  4.  i5a;8  +  sox^  +  45x6  +  51  xS 

+  45x*  +  30x3  +  15x2  +  5x  +  1. 

22.  To  make  x*  +  6x3  +  11  x2  +  ^j.  _|_  5  a  perfect  square,  what  values 
must  be  assigned  to  a  and  h  ? 

Find  the  square  roots  of  the  following  numbers. 

23.  27889.  24.    2313.01.  25.  583.2225. 

26.    4149369.  27.    .00320356.  28.    9.024016. 

Find  approximate  square  roots  of  the  following  numbers  correct  to  the 
third  decimal  figure. 

29.    2.  30.    55.5.  31.    234.561. 

Find  the  cube  roots  of  the  following  numbers. 

32.    1860867.  33.    107284.151.  34.    1036.433728. 


IRRATIONAL   FUNCTIONS  271 

XII.     IRRATIONAL   FUNCTIONS.     RADICALS 
AND    FRACTIONAL    EXPONENTS 

REDUCTION  OF  RADICALS 

Roots.     In  what  follows  the  letters  a,  b,  ■■  •  will  denote posl-    579 
tive  numbers  or  literal  expressions  supposed  to  have  positive 
values. 

Again,  ">/«  will  denote  the  principal  nih.  root  of  a,  that  is, 
the  positive  number  whose  wth  power  is  a  ;  in  other  words,  the 
positive  number  which  is  defined  by  the  formula  (Vo)"  =  a. 

Finally,  when  n  is  odd,  -^  —  a  will  denote  the  principal  nth. 
root  of  —  a,  namely  —  va. 

And  when  we  use  the  word  root  we  shall  mean  principal  root. 

Note.     This  is  a  restricted  use  of  the  word  root;  for  any  number  whose     580 
nth  power  equals  a  is  itself  an  nth  root  of  a,  and  there  are  always  n  such 
numbers,  as  will  be  proved  subsequently. 

Thus,  since  22  =  4  and  (-  2)2  =  4,  both  2  and  -  2  are  square  roots 
of  4.  We  shall  indicate  the  principal  root  2  by  Vi,  the  other  root  -  2 
by  -Vi. 

When  n  is  odd  and  a  is  real,  one  of  the  nth  roots  of  a  is  real  and  of 
the  same  sign  as  a,  and  the  rest  are  imaginary. 

When  n  is  even  and  a  is  positive,  two  of  the  nth  roots  of  a  are  real, 
equal  numerically,  but  of  contrary  sign,  and  the  rest  are  imaginary. 

When  n  is  even  and  a  is  negative,  all  the  nth  roots  of  a  are  imaginary. 

In  the  higher  mathematics  Va  usually  denotes  any  nth  root  of  a,  not, 
as  here,  the  principal  root  only. 

Radicals.     Any  expression  of  the  form  Va  or  b  Va  is  called     581 
a  radical ;  and  n  is  called  the  index,  a  the  radicand,  and  h 
the  coefficient  of  the  radical. 

When  both  a  and  b  are  rational  numbers  or  expressions, 

h  'Va  is  called  a  simple  radical. 

Thus  5  Vi  is  a  simple  radical  whose  index  is  3,  its  radicand  4,  and  its 
coefl&cient  5. 


272  A   COLLEGE   ALGEBRA 

582  Formulas  for  reckoning  with  radicals.  The  rules  for  reckoning 
with  radicals  are  based  on  the  following  formulas,  in  which 
m,  n,  p  denote  positive  integers. 

1.    Va^  =  "-^'aF\ 
2.    ^b  =  Ta.-Vb.  3.    ^~±^^. 

4.    {Vay  =  VoF\  5.    VVa  =  '>/«. 

Observe  in  particular  that,  by  1,  the  value  of  a  radical  is 
not  changed  if  its  index  and  the  exponent  of  its  radicand  are 
multiplied  by  the  same  positive  integer  or  if  any  factor  common 
to  both  is  cancelled ;  thus,  Vo^  =  Vo^.  The  similarity  of  this 
rule  to  the  rule  for  simplifying  a  fraction  is  obvious. 

These  formulas  may  be  proved  by  aid  of  the  definition 
(Va)"  =  a,  the  laws  of  exponents  («.'")"=«""',  (aby  =  a"b", 
and  the  rule  of  equality,  §  261,  3, 

Two  positive  numbers  are  equal  if  any  like  powers  of  these 
numbers  are  equal. 

Thus, 

1.  Vo^  =  Vrt'"^,  since  their  «^;th  powers  are  equal. 
For     (  Va^'yp  =  a"'P  ;  and  (va'")"-^'  =  (a'")''  =  a"v. 

2.  'Vab  =  V a  •  V^,  since  their  nth.  powers  are  equal. 
For  (V^)«  =  ab  ;  and  (V^  ■  %/6)"  =  (V^)"  •  (  v/fe)"  =  ab. 

3    \/-  = >  since  their  nth.  powers  are  equal. 

^^       </b 


(^y  =  ?;andfi^y  =  <-^"=«. 


4.    ( Va)'"  =  Vo^,  since  their  «th  powers  are  equal. 
For         (Va"')»  =  a"';  and  [{Va)">]"  =  [(y/a)"]'»  -  a'". 


IRRATIONAL    FUNCTIONS  273 

5.     V   V  a  =  Va,  since  their  mnth  powers  are  equal. 

For        (  Vfl)"»«  =  a  ;  and  (  V  Va)'"'"  =  ( Va)"  =  a. 
The  following  examples  will  show  the  usefulness  of  these 
formulas. 

1.    -v/s  =  V^  =  V2.  2.    VSab^  =  Vi^  .  V2ab  =  26  V2ab. 


3. 


3/ 3c        Vsc       Vsc         .       5/    ,  10, ; 

V#^  =  I^  =  ^-        '•    ^^32x'V  =  V32x:^.5  =  V2^ 


5.    ( V2^)2  =  V(2  x2/-^)2  =  V4xV  =  V  VT^. 
-  On  simplifying  radicals.     That  form  of  a  radical  is  regarded     583 
as  simplest   in  w^hich   the  radicand   is  the  simplest  integral 
expression  possible.     Hence  for  simplifying  radicals  we  have 
the  following  rules,  which  are  immediate  consequences  of  the 
formulas  just  demonstrated. 

1.  If  the  radicand  he  a  -power  whose  exponent  has  a  factor  in 
common  with  the  index,  cancel  that  factor  in  both  exponent  and 
index. 

Thus,  -^27  x-V  =  \/(3x^  =  v'3xy2. 

2.  //  aiuj  factor  of  the  radicand  be  a  power  ivhose  exponent 
is  divisible  by  the  index,  divide  the  exponent  by  the  index  and 
then  remove  the  factor  from  under  the  radical  sign. 


Thus,  vie x"?/9  =  V2*x'»x37/82/  =  2 xy"^  ^xHj. 

3.  If  the  radicand  he  a  fraction,  multipdy  its  numerator  and 
denominator  by  the  simplest  expression  which  ivill  render  it  pos- 
sible to  remove  the  de^iominator  from  under  the  radical  sign. 

„,  s/xw         3  U xyz        1    ^ n 

Thus,  -V  — ^  =  \\-~  =  7^^^  ^y^- 

'  \2  22        \  8  23        2  2 

Similar  radicals.     Radicals  which,  when   reduced  to  their     584 
simplest  forms,  differ  in  their  coefficients  only  are  said  to  be 
similar. 


Thus,    yf\x^y  and   VsTx^  are  similar ;   for  their  simplest   forms, 
namely  2  x  Vx^  and  9  x^y  "^'^i  differ  in  their  coefficients  only. 


274  A   COLLEGE   ALGEBRA 

585         On  bringing  the  coefficient  of  a  radical  under  the  radical  sign. 

Since  b  V^  =  V'6"a,  the  coefficient  of  a  radical  may  be  brought 
under  the  radical  sign  if  its  exponent  be  multiplied  by  the 
index  of  the  radical. 

EXERCISE  XXXm 

Reduce  each  of  the  following  radicals  to  its  simplest  form. 


1. 

5. 

Vl8.              2. 
V3/2.            6. 
4^25a56iOci5d6. 

V588.                3.    V-272.             4.    V-IOOO. 
V3/2.              7.    V3/4.                8.    \/3/16. 

9. 

10.    Vl28  a26*c8. 
13.    ■s/a"62"c3". 

11.      V8X<52/9215. 

12. 

"^25a26*c6. 

14.    ■v^a2«  +  '63«  +  V». 

15. 

^X6  -  X32/3. 

16.    V(x2- 

2/-)  (•«  +  2/)- 

17. 

18.    ^/a^^>l  - 

-  2  a-'65  +  a--;6«. 

19. 

3  a3  +  63 
\  32  a?)2 

3/        a3 

-■  ^^9w■ 

22. 

,     ,„.3 

-  Vf -^-f -i 

-«*•    Va3«63„  +  2 

Bring  the  coeiiicients  of  the  following  under  the  radical  sign. 


a  -\-b     \a  —  b 
a  -  6 \a  +  b 


25.   3  a  V3  a.  26.    '-^-^^  "V/ '^''  — ,  •        27.  3  ax  %  i  /  27  a3x3 


Show  that  the  following  sets  of  radicals  are  similar. 

28.    V18,   V50,  and   V\J%.  29.    V^,   ^fm,  and  ^8/9. 


30.    V(x3  -  2/3)  (x  _  y)  and  Vx*y2  _i-  xHj^  +  x-y*. 

OPERATIONS  WITH   RADICALS 

586         Addition  and  subtraction.     We  have  the  rule  : 

To  reduce  the  algebraic  sxim  of  two  or  more  radicals  to  its 
simplest  form,  simplify  each  radical  aiid  then  combine  such  of 
them  as  are  similar  by  adding  their  coefficients. 


IRRATIONAL   FUNCTIONS  275 

Example.     Add  Vl6 a^b,  -  Vg a?b,  3  Vi,  and  -  2  Vi/2. 
We  have  VWa^  -  V9a26  +  3  V^  -  2  Vl72. 

=  4aV6-3aV6  +  3V2-V2  =  aV6  +  2V2. 
Observe  that  a  sum  of  two  dissimilar  radicals  cannot  be 
reduced  to  a  single  radical. 

Thus,  we  cannot  have  Vx  +  Vy^  Vx  +  y  except  when  x  or  ?/  is  0 ; 
for  squaring,  we  have  x  +  y  +  2  y/xy  =  x  +  y,  .-.2  Vxy  =  0,  .-.  xy  —  0, 
.-.  either  x  =  0  or  y  =  0. 

Reduction  of  radicals  to  a  common  index.     It  follows  from  the     587 
formula  Va'"  =  Va™^  that  we  can  always  reduce  two  or  more 
radicals  to  equivalent  radicals  having  a  common  index.     The 
least  common  index  is  the  least  common  multiple  of  the  given 
indices. 

Example.     Keduce  va^  and  Vfts  to  their  least  common  index. 

The  least  common  multiple  of  the  given  indices,  6  and  8,  is  24.     And         , 

Va^  =  Va2o  and  VP  -  Vfts. 

Comparison  of  radicals.     We  make  the  reduction  to  a  common     588 
index  when  we  wish  to  compare  given  radicals. 

K. 20^  6  /— 

Example  1.     Compare  vl6,  v6,  and  V3. 

The  least  common  multiple  of  the  given  indices,  15,  10,  6,  is  30  ;  and 

15  , 30  , 30  , 10  ,—  30  , 30  , 6  ,—         30  , 30 , . 

V 16  =  V162  =  V256  ;    V6  =  V63  =  V216;    V3  =  V35  =  V2i3. 
Therefore,  since  256  >  243  >  216,  we  have  V16  >  V3  >  V6. 

Example  2.     Compare  2  V3  and  ViT. 

Bringing  the  coeflBcient  of  the  first  radical  under  the  radical  sign, 
§  585,  and  then  reducing  both  radicals  to  the  common  index  6,  we  have 

2  V3  =  V12  =  v^  =  \/l728  ;   ^/41  =  V^  =  V168I. 
Therefore,  since  1728  >  1681,  we  have  2  Vs  >  ViT. 
Multiplication  and  division.     From  the  formulas  589 

Va  •  V^  =  -y/ab  and 
we  derive  the  following  rule : 


276  A   COLLEGE   ALGEBRA 

To  multiphj  or  divide  one  radical  hy  another,  reduce  them, 
if  necessary  to  radicals  having  the  least  common  index.  Then 
find  the  product  or  quotient  of  their  coefficients  and  radicands 
separately. 

Example  1.     Multiply  4  >/xy  by  2  Vx-y^. 

We  have  4  Vxy  •  2  \/x2^  =  8  V^  •  \/^  =  8  VxV  =  Sxy  Vx?/. 

Example  2.     Divide  6  Vxy  by  2  vxy. 

We  have    6  V^/2  Vxy  =  3  V^/ Vxy  =  3  v^. 

590  Involution.     From  the  formulas 

(V^)'»  =  </a"'  and  'V^"  =  Va"' 
we  derive  the  following  rule  : 

To  raise  a  radical  of  the  form  Va^  to  the  mih  potver,  Cancel 
any  factor  ivhich  may  he  common  to  m  and  the  hidex  of  the 
radical,  and  then  midtijdy  the  exponent  of  the  radicand  by 
the  remaining  factor  of  m. 

Example.     Raise  2  Vxi/2  to  the  9th  power. 
We  have 
(2  -v/^2)9  ^  29  (v^)9  =  128  (V^2)3  =  128  V^  =  128  zy3  Vx. 

591  Evolution.     From  the  formulas 

Vl^tt  =  "V^  and  "Va^  =  ^To™ 
we  derive  the  following  rule : 

To  find  the  mth  root  of  a  radical  of  the  form  Va«,  cancel 
any  factor  which  may  be  common  to  m  and  the  exponent  of  the 
radicand  and  midtiply  the  index  of  the  radical  hy  the  remaining 

factor  of  n, 

i. 

*      Example  1.     Find  the  sixth  root  of  vxV- 

We  have  V  Vx^y*  =  v  y/xy'^  =  Vxy^. 

Example  2.     Find  the  cube  root  of  54  a  Vft. 

We  have     

V^54aV6  =  >y38  •  2  a  Vft  =  3  ^  VT^  =  3  vT^. 


IRRATIONAL   FUNCTIONS  277 

Simple  radical  expressions.     By  a  simjjle  radical  expression     592 
we  shall  mean  any  expression  which  involves  simple  radicals 
only.     Thus,  Va  +  V^  is  a  simple  radical  expression.     We 
call  such  an  expression  integral  when  it  involves  no  fraction 
with  a  radical  in  its  denominator. 

By  the  rules  just  given,  sums,  differences,  products,  and 
jMivers  of  simple  integral  radical  expressions  can  be  reduced 
to  algebraic  sums  of  simple  radicals.  In  §  607  we  shall  show 
that  the  like  is  true  of  quotients.  But  ordinarily  a  root  of  a 
simple  radical  expression,  as  \a  +  V^,  cannot  be  reduced  to 
a  simple  radical  expression. 

Example  1.     Multiply  3V6  +  2V5by2V3-  VlO. 
We  hare 

(3  Ve  +  2  V5)  (2  Vs  -  VlO)  =  6  vTs  +  4  vTs  -  3  Veo  -  2  V50 

=  8  V2  -  2  VT5. 
Example  2.     Square  V2  +  Vi. 
We  have 

(V2  +  \/4)2  =  2  4-2  V2  V^  +  VT6  =  2  +  4^2  +  2-^. 

EXERCISE  XXXIV 

Reduce  the  following  to  their  least  common  index. 

6/-      10/—  IS/—  3, i. 6; 

1.    V3,  V3,  and  V3.  2.    Va^  V2 a^b^  andVTffS. 

Compare  the  following. 

3.    3  Vi  and  2  ^3.  4.    V3,  v^,  and  ^/l. 

Eeduce  each  of  the  following  to  a  simple  radical  in  its  simplest  form. 

5.    V35  -=-  Vfjl.         6.    10  -4-  V5.  7.    4  -  ■\/2. 

8.  Ve  •  VTo  •  VT5.    9.  ^/m  •  ^/m  •  -v/Ts.     lo.  2  V3  •-  3  v^. 

11.    Vi-V^-v^.        12.    -v/3-i-V^.  13.    2V35.  V65H-V91- 

14.    ^a^b^c''  ■  -x/osb^cs.  15.  "'V^  ■  '"v^. 

16.    V^  -  V^.  17.    \/a26c2 .  4/^&2^. 


A   COLLEGE    ALGEBRA 


18.  Va-'v^.  19 

20.  Va62- V'a65^(v'^- Vai26"). 

21.  (Vl2)3.  22.  (\/a2)6. 
24.  Vv^-  25.  ^^/Vs. 
27.  V  v^.  28.    V2V2. 

31.  '•v^V;?'. 


23.  (2Vx2/223)6. 

26.  Vv'a^bVc^. 

29.  V2V2. 

32.  (V    Va)         . 


30.    VV2-.V'2. 

Simplify  each  of  the  following  as  far  as  possible. 

33.    V12  +  V75  -  Vis  +  Vl47.        34.    Vl25  +  VT75  -  V28  +  Vl/20, 

35.    "v^  -  VTo8  +  ^^T/i.  36.    ^^a/hc  +  y^h/ca  +  ^c/ah. 

37.    V50  _  Vii  +  ^-  24  +  ^.       38.    V(^r+6J%  -  V^  -  V62^. 


Vax3  +  6  ax2  +  9  ax  -  Vax^  -  4  a-x^  +  4  a^x. 


40.  (x  +  y)^f—^-{x-y)■ 

41.  (V2  +  V3+V6)-  Ve. 
43.    (V6+V5)(V2  +  Vl6). 
45.   (1+V3)3. 


—  y  \  x2  —  ?/" 

42.  ( V(5  +  VTo  +  VTi) -- V2. 

44.  Vo  +  2  V2  •  V5  -  2  V2. 

46.  (Va  + v'a  +  l)(Va- Va  +  1). 


FRACTIONAL  AND  NEGATIVE  EXPONENTS 

593         In  many  cases  reckoning  with  radicals  is  greatly  facilitated 
by  the  use  oi  fractional  exponeyits. 

Thus  far  we  have  attached  a  meaning  to  the  expression  a" 
only  when  n  denotes  a  positive  integer.  The  rules  for  reckon- 
ing with  such  expressions,  namely, 

1.    a"'  -a"  =  a'"  +  ",  2.    (a"")"  =  a"'",  3.    (ab)"  =  a''b", 

are  among  the  simplest  in  algebra.  It  is  therefore  natural  to 
inquire  :  Can  we  find  useful  meanings  for  a'\  in  agreement 
with  these  rules,  when  n  is  not  a  positive  integer  ? 


IRRATIONAL    FUNCTIONS  279 

The  definition  aP/i  =  Va^.     Tak£_fl.^  f orjnatance.     We  wish,     594 
if  possible,  to  find  a  meaning  for  this  symbol  which  will  be  in 
agreement  with  the  rules  1,  2,  3. 

But,  to  be  in  agreement  with  1,  we  must  have 

(a^y  =  a^.a^  =  J^'^  =  a^  =  a, 

that  is,  a^  must  mean  either   \a  or  —  Va. 

We  choose  the  more  convenient  of  these  two  meanings,  and 
define  a^  as  Va. 

We  thus  find  that  one  of  the  conditions  which  we  wish  a^ 
to  satisfy  suffices  to  fix  its  meaning. 

Similar  reasoning  leads  us  to  define  a}  as   Va,  a'  as  Va^, 

and  in  general  a«  as  '\aP,  that  is,  as  the  principal  qth  root 
of  aP. 

p  pm  p 

Observe  that  since  a''  =  Va**  =  'Vo^  =  a«"',  the  value  of  a« 
is  not  changed  when  ^9/3-  is  replaced  by  an  equivalent  fraction. 
Thus,  a^  —  a^  =  a^\  also  a"^  =  a^  =  a^. 

The  definition  a°  =  1.     Again,  to  be  in  agreement  with  1,     595 
we  must  have 

«<>»"'  =  ftO  +  m^^m^ 

and  therefore  a°  =  «"'/«"'  =  1. 

We  are  therefore  led  to  define  a"  as  1.  596 

The  definition  a~®  =  1/a®.     Finally,  to  be  in  agreement  with 

1,  we  must  have,  §  595, 

ar'-a'  =  a-'  +  '  =  a"  =  1, 

and  therefore  a~*  =  1/a'. 

We  are  therefore  led  to  define  a~^  as  1/a'. 

Thus,  by  definition,  a-s  =  1  /a^,  oT^  =  \/a^  =  \/  V^. 

p 
Jt  remains  to  prove  that  the  meanings  thus  found  for  a',  a°, 

and  a~^  are  in  complete  agreement  with  the  rules  of  exponents. 

Theorem  1 .      The  Iww  a"  a°  =  a'"+°  holds  good  for  all  rational     597 
values  of  m.  and  n. 


280  A   COLLEGE    ALGEBRA 

Let  2h  <1j  ^5  *  denote  any  positive  integers.     Then 

1.  When  m  =p  j q  and  n  =  r/s,  we  have,  §  582, 

-         -  Q I s/ 9*/ 9-'/ 

a'^.a'  =  -^JaP  ■  V  a'"  =  Va^'  •  Va«'" 

qs  / -+- 

2.  When  ??i  =  —  p/q  and  ?i  =  —  r/^,  we  have,  by  Case  1, 

3.  When  m  =p /q  and  n  =■  —  r / s,  and  p/q  >  r / s,  we  have 

4.  When    m=p/q    and   n  =  —  r/s,    and  plq<rlsj    we 
have,  by  Case  3, 

£     _:  1  1  ?+(-^) 


598  Theorem  2.      The  laio  (a"")"  =  a™-"  liolds  good  for  all  rational 

values  of  ni  and  n. 

For,  let  m  denote  any  rational  number.     Then 

1,  When  TO  is  a  p)ositive  integer,  we  have,  §  597, 

(a'")"  =  a™  •  a'"  •  •  •  to  n  factors  =  a'"  +  '"  +  •••""'  "■>■■"«  =  a™. 

2.  When  n  =  p  /  q,  where  ^  and  (7  are  positive  integers,  we 
have,  by  Case  1, 


(a"')9  =  V(a"')"  =  Va"*  =  a  «   =  a      «. 

3.    When  n  =  —  s,  where  s  is  any  positive  rational,  we  have, 
by  Cases  1,  2, 

^     ^  (a"')"       a"" 


IRRATIONAL    FUNCTIONS  281 

Theorem  3.      The  law  (ab)"  =  a"b°  holds  good  for  all  rational     599 
values  of -a. 

1.  Let  7i  =  p/q,  where  p  and  q  denote  positive  integers. 

Then 

p  p  p 

(aby  =  -Vjaby  =  -^a^b"  =  VaJ'  ■  Vbi'  =  a^b^. 

2.  Let   n  =  —  s,    where    s    denotes    any    positive    rational, 
whether  integral  or  fractional.     Then,  by  Case  1, 

^     ^  (aby       a'b" 

Applications.     The   following   examples   will   illustrate  the     600 
use   of    fractional    and    negative   exponents.     A   complicated 
piece  of  reckoning  with  radicals  often  becomes  less  confusing 
when  this  notation  is  empl-^yed. 

Example  1.     Simplify   \  a/^a. 

We  have       ^ a/^a  -.  {aa~^^-  =  («*)i  =  a^  =  Va. 

4, 6. Z. 

Example  2.     Simplify  Va^»  •  \'cv>h  ^  ^a-b^. 

We  have  va^^  •  va'^fe  -^  Va-6^  =  a^b^  ■  a^^  ■  a~h~^ 

Example  3.     Expand  {x^  +  y~^^. 

We  have  (x'  +  y-')^  =  (x^^  +  3(j;3)2y-§  +  3x?(y-i)2  +  (y-^f 

=  X-  +  .3  x'y~  ^  +  3  x%~  ^  +  y~\ 

Example  4.     Divide  x  —  ?/  by  x*  +  x^y^  +  y^. 
Arranging  the  reckoning  as  in  §  401,  we  have 

X  —  y  [x^  +  x'y^  +  y^ 

X  +  x^y^  +  x^y^  I  x»  —  y^ 

-  x^j/^  -  x^2/^  -  y  Hence  the  quotient 

—  x^y^  —  x^2/5  —  y  isz^  —  y^. 


282  A   COLLEGE    ALGEBRA 

EXERCISE  XXXV 

Express  as  simply  as  possible  without  radical  signs 
1.   '-vV.  2.    V^.  3.    aV^. 

4.    bVb*-  V65. 
Express  without  negative  or  fractional  exponents 
5.    a»^.  6.    c-i-5.  7.   ((Z')-6. 

8.    (e-«^)-^. 
Express  with  positive  exponents  and  without  radical  signs 
9.    a-V6-°c-2.  10.    x-^V^.  11.   (l/V^)-4. 

12.    a;-2  Vr-^/?/-2  V^. 
Express  as  simply  as  possible  without  denominators 

13    ^  _  ^  _  a~M^~'  +  c~')  6  +  c 

be      c-2  a-2(6  +  c)  6-    +c-i' 

Reduce  each  of  the  following  to  its  simplest  exponential  form. 
14.   (3i)l  15.    8li.  16.    (-27)1 

17.    8-5.  18.    a?aM.  19.    akr^arA 

20.   (a^6)5a^6l  21.   ab-^/a-^b.  22.    (a')l 

23.   (a-i6-2c3)-2.        24.    (-32aio)l  25.    (- a66-9)-*. 

26.    6-iVF5^6-iVFi.  27.    (a-^  V6^)3. 


28.   (8a-i5/^125a3)-J.  29.    Va^  (fjc- 1)-2. 

30.    Va-iV^.  31.    ■V'a?  V^/V-^^  •  V^. 

32.     [(X*):*]:*.  33.     (x^'  +  a^J';/!''  +  ^)a:  +  ». 

34.  (x' -  r)/(x-i  +  2/--S). 

35.  Multiply  x*  +  x"y^  +  y^  by  x*  -  x^y^  +  y*. 

36.  Divide  a2  _  h^  by  a^  -  6'. 

37.  Expand  (x^  -  yh^)*.  38.    Simplify  [(e^  +  e-^)2  -  4)]*. 

39.  Find  square  root  of  x2  +  4x'y^  +  Axy  -\-  Qx^y^  +  12  2/2  4.  Qx-^yK 

40.  Find  cube  root  of  x^  +  3x2  +  6x  +  7  +  6x-i  +  3x-2  +  xr^. 


IRRATIONAL   FUNCTIONS  283 

THE  BINOMIAL  THEOREM  FOR   NEGATIVE  AND 
FRACTIONAL   EXPONENTS 

If  in  the  binomial  expansion,  §  561, 

(a  +  by  =  a"  +  na"-'b  +  ''^'^~^^  a'^-'b^  +  •  •  • 

we  assign  a  fractional  or  negative  value  to  w,  we  shall  have     601 
on  the  right  a  never-ending,  or  infinite,  series ;  for  none  of  the 
coefficients  n,  n(n  —  l)/2,  •  •  •  will  then  be  0. 

It  will  be  shown  further  on  that  if  b  <  a  the  sum  of  the 
first  m  terms  of  this  series  will  approach  the  value  of  (a  +  by 
as  limit  when  m  is  indefinitely  increased ;  in  other  words,  that, 
by  adding  a  sufficient  number  of  the  terms  of  this  series,  we 
may  obtain  a  result  approximating  as  closely  as  we  please  to 
the  value  of  (a  +  by. 

This  is  what  is  meant  when  iu  is  said  that  the  binomial 
theorem  holds  good  for  (a  +  by  when  n  is  fractional  or  nega- 
tive and  b  <.  a. 

Example  1.     Expand  (8  +  x"')'  to  four  terms. 

Putting  n  =  1/3,  a  =  8,  6  =  x~^  in  the  formula,  we  have 

(8  +  x"')'  =  8s  +  ^  •  8-3X-*  +  5__^8-§(x-^)2 


2 

l(-t)(-D 


8-S(x-^)3  + 


'      5x-? 


12        288      20 


Example  2.     Find  the  sixth  term  in  the  expansion  of  1  /  (a'  -f-  x^)^  or 
(a^  -I-  x^)-2. 

Putting  71  =  —  2,  a  =  a^,  &  =  X3,  r  =  5  in  the  formula  for  the  (r  -|-  l)th 
term,  §  565,  we  have 

(-2)(-3)(-4)(-5)(-6)      .  ,    =  _  6  a-ix^. 

1-2-3-4.6  ^    '  ^    ' 


284  A    COLLEGE   ALGEBRA 

Example  3.     Expand  Vl  +  x  to  four  terms. 

Since  Vl  +  x  =  (1  +  x)^  we  have  n  =  h,  a  =  1,  b  =  x. 

2  2-3 

2       8       16 
The  result  is  the  same  as  that  obtained  in  §  572,  Ex.  1. 
Example  4.     Find  an  approximate  value  of  VlO. 
We  have     VlO  =  (32  +  i)^  =  3  (1  +  i}K 


+   ■ 


6   216   3888 
=  3  +  .16666  -  .00462  +  .00025  +  •  •  •  =  3.1623  nearly 


EXERCISE   XXXVI 

Expand  each  of  the  following  to  four  terms. 
1.    (1  +  x)^.  2.    (a^  +  x'^~K  3.    4^(27  -2x)2. 

4.   (a^  +  x)^.  5.    (a-i-6~5)-4.  q    (^^  +  ^y)-6. 

_  1  o  1 


\  vTXsv^y 


2  +  3x  ^(1  +  x)2  ^  Vl  +3Vx' 

10.  Find  the  tenth  term  in  (1  +  x)-3. 

11.  Find  the  seventh  term  in  (x-2  —  2  ?/*)i 

12.  Find  the  term  involving  x^  in  (1  —  x')». 

13.  Find  the  term  involving  x-2  in  x~^  (2  +  x~s)-3_ 

14.  By  the  method  illustrated  in  §  601,  Ex.  4,  find  approximate  values 
of  the  following. 

1.   Vgo.  2.   V^.  3.   Vsi. 


IRRATIONAL   FUNCTIONS  285 

RATIONALIZING  FACTORS 

Rationalizing  factors.     When  the  product  of  two  given  radi-    602 
cal  expressions  is  rational,  each  of  these  expressions  is  called 
a  rationalizing  factor  of  the  other. 

Thus,  (Va  +  V6)  ( Va  -  Vb)  =  a-h.  Hence  Va  +  Vd  is  a  rational- 
izing factor  of  Va  —  Vft,  and  vice  versa. 

It  can  be  proved  that  every  finite  expression  which  involves 
simple  radicals  only  has  a  rationalizing  factor.  The  following 
sections  will  serve  to  illustrate  this  general  theorem. 

Rationalizing  factors  of  functions  of  square  roots.  Every  expres-  603 
sion  which  is  rational  and  integral  with  respect  to  Va?  can  be 
reduced  to  the  form  A  +  B  V^,  where  A  and  B  are  rational 
and  integral  with  respect  to  x ;  and  A  -]-  B  V^  has,  with 
respect  to  x,  the  rationalizing  factor  A  —  B  V^,  obtained  by 
merely  changing  the  sign  of  -\x. 

Thus,  2(Vx)*  +  3x(Vx)3  may  be  written  2x-  +_3a;2  Vx.  Hence  this 
expression  has  the  rationalizing  factor  2  a;^  _  3  ^1  Vx. 

We  may  obtain  a  rationalizing  factor  of  an  expression 
which  is  rational  and  integral  with  respect  to  any  finite  num- 
ber of  square  roots,  as  Va;,  Vy,  V^,  •  •  ■ ,  by  repetitions  of  the 
process  just  explained.  For  we  shall  obtain  a  result  which 
is  completely  rational  if  we  multiply  the  given  expression 
by  its  rationalizing  factor  with  respect  to  V^,  the  product  by 
its  rationalizing  factor  with  respect  to  Vy,  and  so  on. 

Example.     Find  the  rationalizing  factor  ofl+Vx  +  v^  +  2  Vxy. 

We  have                                   l+Vy  +  \^(l+2  V^).  (1) 

Multiply  (1)  by                        1  +^^-^^(1  +  2  Vy).  (2) 
We  obtain                              (1  +  V^Y  -  x  (1  +  2  V^)% 

or                                      1  -x  +  2/-4xy  +  2V?/{l -2x).  (3) 

Multiply  (3)  by         l-x  +  7/-4xy-2Vy(l-2x).  (4) 

We  obtain                (1  -  x  +  ?/  -  4  xy)"-  -  4  y  (1  -  2  xf.  (5) 

Therefore,  since  (5)  is  completely  rational,  the  product  of  (2)  and  (4) 

is  the  rationalizing  factor  of  (1). 


286  A  COLLEGE   ALGEBRA 

604  Rationalizing   factors   of   binomial   radical    expressions.     The 

rationalizing  lactor  of  an  expression  of  the  form   Va  ±  V6 
may  be  found  as  in  the  following  example. 

Example.     Find  the  rationalizing  factor  of  v a  +  v6. 

We  have  Va  +  Vb  =  a^  +  b^  =  (a"^)^  +  {b^)K  (1) 

But,  §438,   (a2)B  +  (a^)^  will  exactly  divide  the   rational  expression 

a*  -  63,  the  quotient  being  (a^)^  -  (a-')^(65)6  + (53)i.  (2) 

Hence  (2)  is  the  rationalizing  factor  of  (1). 

605  On  rationalizing  the  denominator  of  a  fraction.  Any  irrational 
expression  of  the  form  A  /B,  in  which  B  involves  simple  radi- 
cals only,  may  be  reduced  to  an  equivalent  expression  having 
a  rational  denominator  by  multiplying  both  A  and  B  by  the 
rationalizing  factor  of  B. 

Example  1.     Rationalize  the  denominator  of  1/ Va^. 

TTT    t.  1  1  a*         a^      */- 

We  have  -j —  —  —  =  — — ^  =  —  =  y/a/a. 


V  X-  +  Cl^  +  'Vx- 

Example  2.     Rationalize  the  denominator  of 


Vx2  +  a2  -  Vx2  _  (£1 

We  have 

V^:jr^  +  Vx2  -a?  _  (Vx2  +  a2  +  Vx2-a2)2 


Vx2  +  a2  -  Vx2  -  a2      ( Vx2  +  a2  -  Vx2  -  a2)  ( Vx2  +  02  +  Vx2  -  a2) 


_  x2  +  Vx*  -  a* 
~  ^2 

606  In  computing  an  approximate  value  of  a  fractional  numeri- 
cal expression  which  involves  radicals,  one  should  begin  by 
rationalizing  the  denominator.  Much  unnecessary  reckoning 
is  thus  avoided. 

Example.     Find  an  approximate  value  of  (1  +  Vs)  /  (3  —  Vi)  which  is 
correct  to  the  third  decimal  figure. 

Wehave  i±^  =  il±l^^^lii±^  =  1+ V^  .  2.414- .- . 

3  -  V^       (3  -  V2)  (3  -f-  v^) 


IRRATIONAL    FUNCTIONS  287 

Division  of  radical  expressions.     To  divide  one  radical  expres-     607 
sion  by  another,  we  write  the  quotient  in  the  form  of  a  fraction 
and  then  rationalize  the  denominator  of  this  fraction. 

Example.     Divide  4  +  2V5byl-V2+  V5. 
We  have 

4  +  2  V5  (4 +2V5){1  +  V2-V5)      .     -  3  +  2V2 -V5  +  V1O 


1  4.  V2  +  V5      (1  +  V2  +  V5)  (1  +  V2  -  VS)  V2  -  1 

^  (-3  +  2V2-V5  +  VlO)(V2  +  l)  ^  J  _  V2  +  V^. 
(V2-1)(V2  +  1) 

General  result.  It  follows  from  §  592  and  §  607  that  every 
expression  which  involves  simple  radicals  only  can  be  reduced 
to  an  algebraic  sum  of  simple  radicals. 

EXERCISE    XXXVn 

Find  rationalizing  factors  of  the  following. 

2.    ^^o^VftS.  3.   x?  +  x^  +  x'. 

5.    Vx  +  Vy  +  Vz.  6.    Vxy+Vyz+V^. 

-  Vu.  8.    Vx  +  Vx  +  1. 

10.    ^  -  "^62.  11.    x^  _  yk^ 

13.    1  +  xV'.  14.   X?  +  x^  +  1. 

16.    1  +  V2  +  Vs.  17.    1  +  ^/2. 

19.    v'li +  ^/6+^/3. 

Reduce  each  of  the  following  to  a  fraction  having  a  rational  number 
or  expression  for  its  denominator. 

20.    -J—.  21.    «  +  ^^  -       ^-^ 


1. 

v^. 

4. 

V^  + V6c. 

7. 

y/z  +  y/y  -yfi 

9. 

x^^yK 

12. 

xJ  +  yl 

15. 

3-V5. 

18. 

V9  +  V3  +  I. 

23. 


25. 


V^V62 
1 


6  +  V62  -  a2 

1  +V2 +V3 


I-V2+V3  I+V2  +  V3  +  V6 


288  A  COLLEGE  ALGEBRA 

27.    "^  +  ^^   ■       28.  ^_l_  +  ^^_. 
Vx  +  Vy  +  Vx  +  y  V3  -  1   V3  +  1 

Find  approximate  values  of  tlie  following  expressions  correct  to  the 
third  decimal  figure. 

29.   -A-.  30.    '  +  ^  -       '^^"^ 


V125  V?  V2  +  Vs 


IRRATIONAL   EQUATIONS 

On  solving  an  irrational  equation.  The  general  method  of 
solving  an  irrational  equation  is  described  in  the  following 
rule. 

First,  rationalize  the  equation. 
Next,  solve  the  resulting  rational  equation. 
Finally,  test  all  the  solutions  thus  obtaiiied  in  the  given  equa- 
tion and  reject  those  which  do  not  satisfy  it. 

For,  let  P  =  0  denote  the  given  equation,  and  PR  =  0  the 
rational  equation  obtained  by  multiplying  both  members  of 
P  =  0  by  iZ,  the  rationalizing  factor  of  P.  By  §  341,  the  roots 
of  PR  ^0  are  those  of  P  =  0  and  R  =  0  jointly.  We  dis- 
cover which  of  them  are  the  roots  of  P  =  0  by  testing  them 
in  this  equation. 

Example.     Solve  x  —  7  —  Vx  -  5  =  0. 

Multiplying  both  members  by  the  rationalizing  factor  x  —  7  +  Vx  —  6, 

we  obtain 

(X  _  1)2  _  (X  -  .5)  =  0, 

or  simplifying,  x2-15x  +  54  =  0. 

Solving,  by  §  455,  x  =  9  or  6. 

Substituting  9  for  x  in  x -  7  -  Vx-  5  =  0,  we  have  9 -  7  -  V9-5=0, 
which  is  true.     Hence  9  is  a  root. 

But  substituting  6,  we  have  6  -  7  -  V6  -  5  =  0,  which  is  false.  Hence 
6  is  not  a  root. 

But  observe  that  6  is  a  root  of  the  equation  x  -  7  +  Vx  '-  6  =  0,  obtained 
by  equating  the  rationalizing  factor  to  0  ;  for  6  —  7  +  Ve  —  5  =  0  is  true. 


IRRATIONAL    FUNCTIONS  289 

An  equation  which  involves  the  radical  V  J  may  be  ration-  610 
alized  with  respect  to  this  radical  by  collecting  the  terms 
which  involve  \A  in  one  member  and  the  remaining  terms  in 
the  othei*,  and  then  raising  both  members  to  the  «th  power. 
By  repetitions  of  this  process  an  equation  which  involves 
square  roots  only  may  be  completely  rationalized.  It  follows 
from  §  345  that  this  method  is  equivalent  to  that  described  in 
§  609,  but  it  involves  less  reckoning. 

Example  1.     Solve  V  Vx  +  a  =  Vft. 

Cubing  both  members,  Vx  +  a  =  6^. 

Transposing  and  squaring,  x  =  (IP  —  a)2. 

Substituting  this  result  in  the  given  equation,  we -find  it  to  be  a  root. 

Example  2.     Solve 

Transposing,  Vx  —  4 
Squaring,  x  —  4  =  81 

Simplifying,  vx  +  5  =  5. 
Squaring,  x  +  5  =  25. 

Solving,  X  =  20. 

Substituting  20  for  x  in  the  given  equation,  we  have  V25  +  Vi6  =  9, 
which  is  true.     Hence  20  is  a  root. 

Notes.     1.    Observe,  as  in  Ex.  1,  that  we  rationalize  an  equation  with      611 
respect  to  the  unknown  letter  only  and  make  no  attempt  to  rid  it  of  radicals 
which  do  not  involve  this  letter. 

2.  Observe  also  that  an  irrational  equation  may  have  no  root. 

Thus,  the  equation  Vx  +  5  —  Vx  —  4  =  1)  has  no  root.  For  if  we 
attempt  to  solve  it  we  shall  merely  repeat  the  reckoning  in  Ex.  2  and 
shall  again  obtain  the  result  x  =  20  ;  and  V25  —  Vl6  =  9  is  false. 

3.  We  may  add  that  the  simplest  method  of  rationalizing  an  equation 
of  the  form  Va  +  V^  +  Vc  +  Vb  =  0  (or  Va  +  Vb  +  Vc  +  E  =  0) 
is  to  begin  by  writing  it  thus  : 

Va+Vb  =  -Vc  -Vd  (otVa  -\-Vb  =  ~Vc  -E) 

and  then  to  square  both  members.     The  resulting  equation  will  involve 
but  two  radicals  and  it  may  be  rationalized  as  in  Ex.  2. 


290  A   COLLEGE    ALGEBRA 

612  Simultaneous  irrational  equations.  To  solve  a  system  of  such 
equations  we  may  first  rationalize  each  equation,  then  solve 
the  resulting  rational  system,  and  finally  test  the  results  thus 
obtained  in  the  given  system. 

But  if  the  equations  are  of  the  form  described  in  §  379,  they 
should  be  solpd  by  the  method  there  explained. 


Exampk 


soiFea  D} 
1.  I  Solve 


5  +  V^  +  5  =  Vx  +  V^, 
x  +  2y=\-i. 


Squaring  (1),  x-5+?/  +  5  +  2  Vxy  +  ox-  by 


•Zo  ■=  X  -\-  y  ■]- 2  Vary, 

or  y/xy  ■\-  bx  -  by  - -lb  =  Vxy.  (3) 

Squaring  (3)  and  simplifying,      x  —  y  =  b.  (4) 

Solving  (4),  (2),  x  =  9,  y  =  4.  (5) 

Substituting  x  =  9,  ?/  =  4  in  (1),  we  have  Vi  +  V9  =  V9  +  Vi,  whicli 

is  true.     Hence  x  =  9,  ?/  =  4  is  tlie  solution  of  (1),  (2). 

Example  2.     Solve     Vx  +  6  +  2/Vy  =  4,  (1) 

2  Vx  +  6  +  6/%/?/ =  9.  (2) 

Solving  for  Vx  +  6  and  1  /Vy,  we  find  Vx  +  6  =  3,  1  /Vy  =  1  /2.    (3) 
And  from  (3)  we  obtain  x  =  3,  2/  =  4,  which  is  the  solution  of  (1) 
and  (2). 

EXERCISE   XXXVm 

Solve  the  following  equations  for  x. 
1.    x*  =  4.  2.    x-^  =  3.  3.    x?  =  8. 

V3. 

d. 

--7 


4.   {V2x-  1)^ 

6.      Vox  +  Vftx  +  Vex  : 

8.    Vx  +  4  +  Vx  -t  11 


V  ^  +  VS  +  Vx  =  2. 


7.    V4x2  +  x+  10  =  2x+l. 

9.    V4X  +  5+ Vx+1-  V9x+10=0. 


10. 
12. 
13. 


'x  +  1  + 


=  0. 


Vx  +  2 
/x  +  V  +  Vx  -  2  =  Vx  +  2  +  Vx  -  1. 


11.      Vx2  +  3x  -  1  -  Vx'-^-X-  1: 


'x  + 


3  + Vx 


=  2. 


14. 


J. 

^x+1 


'    +    ' 


^X-1         Vx2-1 


IRRATIONAL   FUNCTIONS  291 

Solve  the  following  for  x  and  y. 


Vx  +  17  +  Viy  -  2  =  Vx  +  5  + 


16,  r_^-     __  3, 

+  2  V  X  +  2/  -  4 

17.    Show  that  Vx  +  a  +  Vx  +  6  +  Vx  +  c  +  Vx  +  d  =  0  will  reduce 
to  a  rational  equation  of  the  first  degree. 


18.    Show  that  Vax  +  h  +  vcx  +  d  -  Vex  +/  =  0  will   reduce  to  a 
rational  equation  of  the  first  degree  if  Va  +  Vc  —  Ve  =  0. 


QUADRATIC   SURDS 

Surds.     Numerical  radicals  like  V2'and  Vd,  in  Avhich  the     613 
radicaud  is  rational  but  the  radical  itself  is  irrational,  are  called 
surds.     A  surd  is  called  quadratic,  cubic,  and  so  on,  according 
as  its  index  is  two,  three,  and  so  on. 

Theorem  1.      The  product  of  two  dissimilar  quadratic  surds     614 
is  a  quadratic  surd. 

Suppose  that  when  the  surds  have  been  reduced  to  their 
simplest  forms,  their  radical  factors  are  Va  and  Vi,  The 
product  of  va  and  V ^  is  Va6,  and  this  is  a  surd  unless  ah  is 
a  perfect  square. 

But  ab  cannot  be  a  perfect  square,  since  by  hypothesis  a 
and  b  are  integers  none  of  whose  factors  are  square  numbers, 
and  at  least  one  of  the  factors  of  a  is  different  from  every 
factor  of  b. 

Thus,  V2  ■  Vs  =  Ve,  Ve  •  Vis  =  V90  =  3  Vio. 

Theorem  2.     The  sum  and  the  difference  of  two  unequal  quad-     615 
ratic  surds  are  irrational  numbers. 

This  is  obvious  when  the  surds  are  similar. 
Hence  let  Va  and  V^  denote  dissimilar  surds. 


292  A   COLLEGE   ALGEBRA 

Suppose,  if  possible,  that  Va  +  V^  =  c,  (1) 

where  c  is  rational. 

Squaring  both  members  of  (1)  and  transposing, 

2  Vo^  =  0^  -a-h,  (2) 

which  is  impossible  since  2  Vo^  is  irrational,  §  614,  while 
c^  —  a  —  h  is  rational. 

Theorem  3.  i/"  a  +  Vb  =  c  +  Vd,  ivhere  Vb  and  Vd  are 
surds,  then  a  =  c  and  b  —  (\.. 

For,  by  hypothesis,  V^  —  V(/  =  c  —  a. 

But  this  is  impossible  unless  Vft  —  V^  =  0  and  c  —  a  =  0, 
since  otherwise  V^  —  Vc^  would  be  irrational,  §  615,  and  equal 
to  c  —  a,  which  is  rational. 

Hence  b  =  d  and  a  =  c. 

Square  roots  of  binomial  surds.     We  have 

(  V^  ±  V?/)2  =  x  +  ij±2  Vxy. 

Hence  if  a  +  2  V^  denote  a  given  binomial  surd,  and  we  can 
find  two  jMsiti a e  rational  numbers  x  and  3/  such  that 

X  -\-  1/  =  a  and  cry  =  b, 

then  Vo;  +  Vy  will  be  a  square  root  of  a  +  2  V^  and  Vx  —  Vy 
will  be  a  square  root  oi  a  —  2  Va,  and  both  these  square  roots 
will  be  binomial  surds. 

When  such  numbers  x,  y  exist  they  may  be  found  by 
inspection. 

Example  1.     Find  the  square  root  of  37  —  20  V3. 

Reducing  to  the  form    a  -  2  %^,    37  -  20  \^  =  37  -  2  VSOO. 

But  300  =  25-12  and  37  =  25  +  12. 


Hence  V37  -2  V3OO  =  V'25  -  Vl2  =  5  -  2  V3. 

Example  2.  Find  the  square  root  of  13/12  +  V5/6. 

_    ,                         13          \l      13  ,    V30      13  +  2V3O 
We  have  12  +  Vc  =^  I^  +  "^  =  — 1^ 


IRRATIONAL    FUNCTIONS  29S 

Since  30  =  10  •  3  and  13  =  10  +  3,  we  have  Vi3  +  2  Vio  =  VlO  +  V3. 


Hence 
Note.     We 

may 
sis 

(4), 

^13  +  2  V30 
12 

obtain  formv 

Vx  +  v^  = 

V-^-V-y. 

)y  (2),  x-y-- 

x  +  y- 

X  - 

V10  +  V3      > 
V12 
lias  for  X  and  y  ; 

/12O+V36       1 
12                2 

as  follows  : 

a  —  Va2  _ 
^~             2 

V30 

By  hypothe 

=  Va  +  2  A 

(1) 

and 

=  Va-2V6. 

(2) 

Multiplying 
But 

=  Va2-45. 
=  a. 

(3) 

(4) 

Solving  (3), 

a  +  Va'-  -4b 
2 

-45 

Observe  that  these  values  are  rational  only  v^hen  a^  -  4  6  is  a  perfect 
square.  Hence  in  this  case  only  is  the  square  root  of  a  +  2  Vft  a  bino- 
mial surd. 

EXERCISE   XXXIX 

Find  square  roots  of  the  following. 

1.9  +  V56.  2.    20  +  2  V96.  3.    32-2  Vl75. 

4.    1+- — -.  5.    7-3V5.  6.    8V2  +  2V3O. 

5 

7.    2  (a  +  Va2  -  62).  8.    6  -  2  Vab  -  a^. 

Simplify  the  following. 


618 


9.  Vn  +  12  V2.  10.  V9  +  4  V4  +  2  V3. 

IMAGINARY  AND   COMPLEX  NUMBERS 

Complex  numbers.     Since  all  even  powers  of  negative  num-     619 
bers  are  positive,  no  even  root  of  a  negative  number  can  be  a 
real  number.     Such  roots  are  hnag'mary  numbers. 

Definitions  of  the  imaginary  numbers  and  of  the  operations 
by  "which  they  may  be  combined  are  given  in  §§  217-228,  which 
the  student  shoiild  read  in  this  connection. 

According  to  these  definitions, 

1.    The  symbol  i  =  V—  1  is  called  the  unit  of  imaginaries. 


294  A   COLLEGE    ALGEBRA 

2.  Symbols  of  the  form  at,  where  a  is  real,  are  called  pure 
imaginaries. 

3.  Symbols  of  the  form  a  +  bi,  where  a  and  b  are  real,  are 
called  complex  numbers. 

4.  Two  complex  numbers  are  equal  when,  and  only  when, 
their  real  parts  and  their  imaginary  parts  are  equal,  so  that 

If  a  -\-  bi  =  c  -^  di,  then  a  =  c  and  b  =  d. 

5.  The  siun,  difference,  jyroduct,  or  quotient  of  two  complex 
numbers  is  itself  a  complex  number  (in  special  cases  a  real 
number  or  a  pure  imaginary)  which  may  be  found  by  applying 
the  ordinary  rules  of  reckoning  and  the  relation  i^  =  —  1.  The 
like  is  true  of  any  positive  integral  power  of  a  complex  number, 
since  by  definition  (a  +  biy  =  (a.  +  bi)  (a  +  bi)  ■  ■  ■  to  n  factors. 

Example  1.     Add  5  +  3  i  and  2  —  4  i. 

We  have  5  +  3  i  +  (2  -  4  i)  =  (5  +  2)  +  (3  -  4)  i  =  7  -  i. 

Example  2.     Subtract  6  +  2  i  from  3  +  2  i. 

We  have  3  +  2  i  -  (6  +  2  i)  =  (3  -  6)  +  (2  -  2)  i  =  -  3. 

Example  3.     Multiply  2  +  3  i  by  1  +  4  i. 

We  have  (2  +  3  i)  (1  +  4  i)  =  2  +  3  i  +  8  i  +  12  ^2 

=  2  +  3i  +  8i-12  =  -  10+  lit. 
Example  4.     Expand  (1  +  i)2. 
We  have    (1  +  i)2  =  1  +  2  i  +  i2  =  1  +  2  i  -  1  =  2  i. 
Example  5.     Find  real  values  of  x,  y  satisfying  the  equation 

(x  +  yi)  i  -  2  +  4  i  =  (x  -  2/x)  (1  +  i). 
Carrying  out  the  indicated  operations,  we  have 

-  (y  +  2)  +  (X  +  4) i  =  (X  +  2/)  +  (X  -  2/)i. 
Equating  the  real  and  the  imaginary  parts,  §  619,  4, 
-  (y  +  2)  =  X  +  y  and  x  +  4  =  x  -  y, 
or,  solving,  x  =  6,  y  =  —  4. 

In  §§  238-241  we  have  given  a  method  for  representing 
complex  numbers  by  points  called  their  graphs,  and  rules  for 
obtaining  from  the  graphs  of  two  complex  numbers  the  graphs 


IRRATIONAL    FUNCTIONS  295 

of  their  sum  and  product.     Let  the  student  apply  these  rules 
to  Exs.  1,  3,  4. 

Conjugate  imaginaries.     Two   complex   numbers  like  a  +  hi     620 
and  a  —  bi,  which  differ  only  in  the  signs  connecting  their  real 
and  imaginary  parts,  are  called  conjugate  imaginaries. 

The  product  of  two  conjugate  imaginaries  is  a  positive  real     621 
number. 

Thus,  (a  +  hi)  (a  -  hi)  =  a^  -  hH'^  =  a2  +  62, 

Hence  a  fraction,  as  (a  +  hi)  /  (c  +  di),  may  be  reduced  to    622 
the  form  of  a  complex  number  by  multiplying  both  its  terms 
by  the  conjugate  of  its  denominator. 
Example.     Divide  54-7iby2  —  4z. 
We  have  M^jj  ^  (5  +  7^)  (2  +  4  Q 

2  -  4  i      (2  -  4  i)  (2  +  4  i) 

-  18  +  34  i  _  _  ^       17  , 


20  10      10 

The  powers  of  i.    From  the  equation  p  =  —  1  it  follows  that     623 

the  even  powers  of  i  are  either  —  1  or  1,  and  the  odd  powers 

either  i  or  —  i. 

Thus,  i3  =  z2 .  i  =  —  i ;  i<  =  i3  .  ^  _  _  i .  j  —  _  ^2  _  j  .  ^^^  go  on. 

To  find  the  value  of  i"  for  any  given  value  of  n,  divide  n  by  4.     Then, 
according  as  the  remainder  is  0,  1,  2,  3,  the  value  of  i«  is  1,  i,  —  1,  —  i. 
Thus,  i'^  —  (i*)8  =  1  ;  i^  =  i-^  ■  i  =  i;  and  so  on. 

Even  roots  of  negative  numbers.     The  number  —  4  has  the     624 
two  square  roots  2i  and  —2i;  for   (2iy  =  2^i^  =  —  4,  and 
(—  2iy  =  (—  2y^i^  =  —  4.     We  select  2  i  as  the  principal  square 
root,  and  write  V—  4  =  2i  and  —  V—  4  =  —  2  i. 

Similarly  the  principal  square  root  of  any  given  negative 
number  —  a  is  Vai,  that  is,  V—  a  =  ^/ai. 

From  this  definition  of  principal  square  root  it  follows  that     625 
it  —  a  and  —  b  are  any  two  negative  numbers,  then 

V—  a  V—  b  —  —  -Vab. 
For  V-a  •  V-  6  =  Vai  •  Vbi  =  i^VaVb  =  -  Vab. 


29G  A   COLLEGE    ALGEBRA 

Thus,  while  the  product  of  the  principal  square  roots  of  two 
negative  numbers  —  a,  —  b,  is  07ie  of  the  square  roots  of  their 
product  ab,  it  is  not,  as  in  the  case  of  real  numbers,  the  principal 
square  root  of  this  product. 

When  reckoning  with  imaginaries,  it  is  impo£tant  to  bear  in 
mind  this  modification  of  the  rule  Va  Vb  =  ^(ib.  All  chance 
of  confusion  is  avoided  if  at  the  outset  we  replace  every 
symbol   V—  a  by   Vai. 

Example  1.     Simplify  V^  •  ( V^)5  •  (^^5)7. 

We  have  V^  .  ( V^)5 .  ( V^)"  =  V2  i  •  ( Vs  j)M^  0^ 

=  V2-  (V3)5-  {V5)Tii3 

=  1125  VsOi. 
Example  2.     Multiply  2  +  V^  by  1  +  V^l. 
We  have        (2  +  V~9)  (i  +  y^^)  =  (2  +  3  /)  (1  +  i)  =  -  1  +  5  i. 

626  The  higher  even   roots  of  negative  numbers  are  complex 
numbers.     This  will  be  proved  subsequently. 

Thus,  one  of  the  fourth  roots  of  —  4  is  1  +  i ;  for 
(1  +  j)4  =  1  +  4  i  +  G  i^  +  4  (3  +  i4  =  1  +  4  i  _  6  -  4  i  +  1  =  -  4. 

627  Square  roots  of  complex  numbers.     As  will  be  proved  farther 
■  on,  all  roots  of  complex  numbers  are  themselves  complex  num- 
bers.    We  may  find  their  square  roots  as  follows. 

We  have  (v^  ±  i  Vy)- =  x  —  y  ±  2  t  Vxy. 

Hence,  if  a  +  6J  denote  a  given  complex  number  in  which  h  is  positive, 
and  we  can  find  two  positive  numbers  x  and  y  such  that 

X  -  y  =  a,     (1)      and      2  Vxy  =  b,     (2) 

then  Vx  +  t  Vy  will  be  a  square  root  of  a  +  bi,  and  Vx  —  iVy  will  be  a 
square  root  of  a  —  bi. 

We  may  find  such  numbers  x  and  y  as  follows. 

By  hypothesis  Vx  +  iVy  =  Va  +  bi,  (3) 

and  Vx  -  iV^  =  Va  -  bi.  (4) 

Multiplying  (3)  by  (4),  x  +  y  =  VoM^.  (6) 


IRRATIONAL   FUNCTIONS 


But,  by  (1),  x-y  ■ 

Hence,  solving  (5)  and  (6),  x 


297 
(6) 


Va2  +  62 


and 


a+Va2  + 62 


And  both  these  values  are  positive  since  V a2  +  62  >  a. 

Example.     Find  a  square  root  of  —  1  +  4  Vs  t. 

Here  a  =  -l  and  Va2  +  62  =  V(_  1)2  +  (4  y/l)i 


Hence                    x  =  (-l+9)/2  = 

4  and  2/  =  (1  +  9)/2  =  5. 

Therefore                 V_i+4 

V5i- 

2  +  V5 ». 

EXERCISE 

XL 

Simplify  the  follovring. 

1.    V-49.           2.    V-  18. 

3. 

V-8-V-12.        4.    V-22. 

5.    (V^)2.         6.    ii2. 

7. 

i-'.                          8.    ii5. 

9,   y/x-y-y/y  -X. 

10.    (2+V-3){l+V-2). 

11.    (V-2)7(V-3)9. 

12.    (l  +  2z)3  +  (l  -2iY. 

!'■  A—  %• 

.       V-a2      iV62 

14.    4  +  6i      4-6i^ 
1+i         1-i 

15.   ( V3  +  4  i  +  V3  -  4  i)2. 

16.     (l  +  i3)/(l+i). 

,^    a  +  6i 

9  +  3V2i 

a-6i  (3  +  V2i)(l +V2i) 

19.  Divide  4  by  1  +  V3^. 

20.  Find  a  fourth  root  of  -  16. 

21.  Show  that  (-  1  +  V3i)/2  is  a  cube  root  of  1. 

22.  Show  that  (1  +  i)/V2  is  a  fourth  root  of  -  1. 

23.  Find  real  values  of  x  and  y  satisfying  the  equation 

3  +  2  i  +  X  (i  -  1)  +  2  2/i  =  (3  i  +  4)  (x  +  J/). 

Find  square  roots  of  the  following. 

24.  5  +  12t.  25.   2i  26.    4a6  +  2(o2  -  68)i 


298  A   COLLEGE   ALGEBRA 

XIII.     QUADRATIC    EQUATIONS 

628  General  form  of  the  equation.  Every  quadratic  equation  in 
one  unknown  letter,  as  x,  may  be  reduced  to  the  form 

ax?  +  hx  +  c  =  0, 
where  a,  h,  and  c  denote  known  numbers. 

If,  as  may  happen,  6  =  0,  the  equation  is  called  a  pxire 
quadratic ;  if  &  ^  0,  it  is  called  an  affected  quadratic. 

629  The  roots  found  by  inspection.  The  roots  of  the  equation 
ax^  -^  bx  -\-  c  =  0  are  those  particular  values  of  x  for  which 
the  polynomial  ax'^  -\- hx  +  c  vanishes,  §  332.  There  are  two 
of  these  roots. 

If  the  factors  of  ax^  +  bx  -{-  c  are  known,  the  roots  of 
ax^  +  bx  -{-  c  =  0  are  also  known,  for  they  are  the  values  of  x 
for  which  the  factors  of  ax"  +  bx  -\-  c  vanish,  §§  253,  341.  If 
the  factors  are  x  —  a  and  x  —  /3,  the  roots  are  a  and  /3. 

Example  1.     Solve  the  equation  a;^  _|_  ^  —  6  =  0. 

We  have  x2  +  x  -  6  =  (x  +  3)  (x  -  2). 

The  factor  x  +  3  vanishes  when  x  =  —  3,  and  the  factor  x  —  2  vanishes 
when  X  =  2.     Hence  the  roots  are  —  3  and  2. 
.  Example  2.     Solve  abz^  —  (a^  +  b'^)  x  +  (a^  —  b-)  =  0. 

Factoring,  by  §  443,  [ax  -  (a  +  b)]  [bx  -  {a  -  b)]  =  0. 

Hence  the  roots  are  (a  +  b)/a  and  (a  —  b)/b. 

In  particular,  since  x'^  —  q  =(x  —  Vy)  (x  +  V^),  the  roots  of 
the  pure  quadratic  x^  —  q  =0  are  Vy  and  —  Vy. 

Again,  since  ax^  -i- bx  =  (ax  +  b)x,  the  roots  of  a  quadratic 
of  the  form  ax^  -\-  bx  =  0  are  —b  I  a  and  0. 

Thus,  the  roots  of  4  x2  =  9  are  3 /2  and  -  3 /2  ;  the  roots  of  2  x2  -  x  =  0 
are  0  and  1  /2  ;  the  roots  of  5x2  =  0  are  0  and  0. 

630  Conversely,  to  obtain  tlie  quadratic  whose  roots  are  two 
given  numbers,  as  a  and  y8,  we  form  the  product  (.r  —  a)  (x  —  ft) 
and  equate  this  product  to  0. 

Thus,  the  quadratic  whose  roots  are  -  2  and  1  /3  is  (x  +  2)  (x  -  1/3)  =  0, 
or3x2  +  6x-2  =  0. 


QUADRATIC    EQUATIONS  299 

Example  1.     Solve  the  following  quadratics. 

1.    x2  +  2x-8  =  0.  2.    2x2-7x  +  3  =  0. 

3.    (2x-l)(x-2)  =  x2  +  2.  4.    (x-l)(x-3)  =  {2x-l)2. 

Example  2.     Find  the  quadratics  whose  roots  are 

1.    -2/3,-3/2.  2.  a,  -a.  3.1/4,0. 

General  formula  for  the  roots.     But  ax^  -\-  bx  +  c  may  always     631 
be  factored ;  for,  as  was  shown  in  §  444, 
■    ax^  -}-  bx  +  c 

^  a  [x  -  -  ^>  +  ^^'  -  4  acl  f      _  -b--^b^-A  acl 

Therefore,  since  the  roots  of 

ax""  +  bx  +  c  =  0  (1) 

are   the  values   of  x  for  which  the  factors   of  ax^  -\-  bx  -{-  c 
vanish,  these  roots  are 


-b  +  Wb^-4ac        ,  _  J  _  V^»2  -  4  ac 

X  — — ■ and  X  = — — — - — ■  j 

2  a  2  a 

or,  as  we  usually  write  them, 


-b±  V^/2  _  4  ac 

X  = (2) 

2  a  ^  '^ 

This  formula  (2)  should  be  carefully  memorized,  for  ,it 
enables  one  by  mere  substitution  to  obtain  the  roots  of  any 
given  quadratic  which  has  been  reduced  to  the  form  (1). 

Example.     Solve  4  x2  +  105  x  =  81. 

Reducing  to  the  form  (1),  4x2  +  105 x  -  81  =  0. 

•     Here  a  =  4,  6  =  105,  and  c  =  - 81. 

-  105  ±  V1052  +  4-4.81     ,,    ,  .      3 

Hence      x  = = »  that  is,  -  or  -  27. 

8  4 

When  b  is  an  even  integer,  a  and  c  also  being  integers,  it  is     632 

more  convenient  to  use  the  formula 

-b/2±^(b/2r-ac 
^   x  = ,  (3) 

which  is  obtained  by  dividing  the  numerator  and  denominator 
of  (2)  by  2. 


300  A   COLLEGE   ALGEBRA 

Example.     Solve  3  x^  +  56  x  -  220  =  0. 

Here  6/2  =  28,  and  substituting  in  (3)  we  have 


-  28  ±  V282  +  3.220     ^,    ,  .      10 

x=  = )  that  IS,  —  or  -22. 

3  3 

633  Any  given  quadratic  may  also  be  solved  by  applying  directly 
to  it  the  process  of  completing  the  square,  §  444,  as  in  the 
following  example.  But  since  this  method  involves  needless 
reckoning,  it  is  not  to  be  recommended  —  except  when  the 
formula  of  §  631  has  been  forgotten. 

Example.     Solve  3i2-6x  +  2  =  0. 

Transposing  the  known  term  and  dividing  by  the  coeflBcient  of  x^, 

x2-2x  =  -2/3. 
Completing  the  square  of  the  first  member, 

x2-2x  + 1  =  1/3. 
Extracting  the  square  root  of  both  members, 

a;  _  1  =  ±  V3/3,  whence  x  =  (3  ±  V3)/8. 

634  The  methods  just  explained  enable  one  to  solve  oxij  fractional 
equation  which  yields  a  quadratic  when  cleared  of  fractions. 
But  see  §§  524-527. 

Example  1.     Solve 1 = ^ • 

x+1      x+2      x+3      x+4 

Clearing  of  fractions  and  simplifying,     2  x^  +  10  x  +  11  =  0. 

c  ,   .                                                         _5±V3 
Solvmg,  x  = 

Both  of  these  values  of  x  are  roots  of  the  given  equation,  for  they 
cause  none  of  its  denominators  to  vanish. 

Example  2.     Solve  ^^^  +  ^^  +  ""^  =  0. 

^  X2-1        X2-X        X2  +  X 

Clearing  of  fractions  by  multiplying  by  the  lowest  common  denominator 
x(x2  —  1),  and  simplifying,  we  obtain 

3x2  +  2x-5  =  0,  whence  x  =  1  or  -  5/3. 

But  1  cannot  be  a  root  of  the  given  equation,  since  its  first  two 
denominators  vanish  when  x  =  1.  Hence  —  5/3  is  the  only  root  of  this 
equation. 


QUADRATIC    EQUATIONS  301 


EXERCISE  XLI 

Solve  the  following  equations. 

1.    a;2  +  2x  =  35. 

3.   x2  =  10  X  -  18. 

5.    2x2  + 3x- 4  =  0. 

7.   x2  +  9x- 252  =  0. 

9.    8x2 -82x  + 207  =  0. 
11.    x2-3x-  1  -f  V3  =  0. 

13.  (X  -  2)2(x  _  7)  =  (X  +  2)  (X  -  3)  (x  -  6). 

14.  ^:^  +  ^+-2  =  2. 
X  +  2        2x 

Q  ^ 

16. 


2. 

4x2-4x=3. 

4. 

9x2  +  6x  +  5  =  0. 

6. 

(2x-3)2  =  8x. 

8. 

12  x2  +  56x- 255  =  0. 

10. 

15x2 -80x- 64  =  0. 

12. 

x2-(6  +  i)x  +  8  +  2i  =  0. 

?)(x 

-6). 

15. 

X                 x-\ 

17 

3               1             2x 

2(x2-l)      4x  +  4      8  2x  +  l      4x-2      1-4x2      8 

18.    2x-1^3x+^^5x-U_       ^g  _^+i L_  +  i.  =  o. 

x-2       x-3        x-4  x(x-2)      2x-2      2z 

20.    -A ^  =  ^_._^_. 

X  —  1      4  —  X      X  —  2      3-x 

2^  x  +  3  ,  2x  +  l  17a;  +  7  ,„ 


4(x  +  2)(3x-l)      3(3x-l)(x  +  4)      6(x  +  4)(x  +  2) 

22.  ^  +  ^         + ^— +        ^  +  ^        =0. 

2  x2  -  7  X  +  3      x2  -  2  a:  -  3      2  x2  +  a;  -  1 

23.  3  x2  +  (9  a  -  1)  X  -  3  a  =  0.       24.   x2  -  2  ax  +  a^  -  6^  =  0. 
25.    c2x2  +  c  (a  -  6)  X  -  a6  =  0.        26.    x2  -  4  ax  +  4  a2  -  62  =  q. 

27.  x2  -  6 acx  +  a2(9 c2  -  4 62)  =  o. 

28.  (a2  -  62)a;2  -  2  (a2  +  62)^  +  a2  _  52  =  0. 

29.  l/(x-a)  + l/(x-6)  + l/(x-c)  =  0. 

30     {x-aY-{x-W        ^^^^    ^^ 
(X  -  a)  (x  -  6)         a2  -  W- 


< 


302  A    COLLEGE   ALGEBRA 

EXERCISE  XLU 

1.  Find  two  consecutive  integers  whose  product  is  506. 

2.  Find  two  consecutive  integers  the  sum  of  whose  squares  is  481. 

3.  Find  two  consecutive  integers  the  difference  of  whose  cubes  is  91. 

4.  Find  three  consecutive  integers  the  sum  of  whose  products  by 
pairs  is  587. 

5.  Find  a  number  of  two  digits  from  the  following  data  :  the  product 
of  the  digits  is  48,  and  if  the  digits  be  interchanged  the  number  is 
diminished  by  18. 

6.  The  numerator  of  a  certain  fraction  exceeds  its  denominator  by  2, 
and  the  fraction  itself  exceeds  its  reciprocal  by  24/35.    Find  the  fraction. 

7.  A  cattle  dealer  bought  a  certain  number  of  steers  for  $1260. 
Having  lost  4  of  them,  he  sold  the  rest  for  $10  a  head  more  than  they 
cost  him,  and  made  $260  by  the  entire  transaction.  How  many  steers 
did  he  buy  ? 

8.  A  man  sold  some  goods  for  $48,  and  his  gain  per  cent  was  equal  to 
one  half  the  cost  of  the  goods  in  dollars.     "What  was  the  cost  of  the  goods  ? 

9.  If  $4000  amounts  to  $4410  when  put  at  compound  interest  for  two 
years,  interest  being  compounded  annually,  what  is  the  rate  of  interest  ? 

10.  A  man  inherits  $25,000,  but  after  a  certain  percentage  has  been 
deducted  for  the  inheritance  tax  and  then  a  percentage  for  fees  at  a  rate 
one  greater  than  that  of  the  inheritance  tax,  he  receives  only  $22,800. 
"What  is  the  rate  of  the  inheritance  tax  ? 

11.  A  man  bought  a  certain  number  of  $50  shares  for  $4500  when  they 
were  at  a  certain  discount.  Later  he  sold  all  but  10  of  them  for  $5850 
when  the  premium  was  three  times  the  discount  at  which  he  bought  them. 
How  many  shares  did  he  buy  ? 

12.  The  circumference  of  a  hind  wheel  of  a  wagon  exceeds  that  of  a 
fore  wheel  by  8  inches,  and  in  traveling  1  mile  this  wheel  makes  88  less 
revolutions  than  a  fore  wheel.     Find  the  circumference  of  each  wheel. 

13.  A  square  is  surrounded  by  a  border  whose  width  lacks  1  inch  of 
being  one  fourth  of  the  length  of  a  side  of  the  square,  and  whose  area  in 
square  inches  exceeds  the  length  of  the  perimeter  of  the  square  in  inches 
by  64.     Find  the  area  of  the  square  and  that  of  the  border. 


QUADRATIC    EQUATIONS  303 

14.  The  corners  of  a  square  the  length  of  whose  side  is  2  are  cut  off 
in  sucli  a  way  that  a  regular  octagon  remains.  What  is  the  length  of  a 
side  of  this  octagon  ? 

15.  A  vintner  draws  a  certain  quantity  of  wine  from  a  full  cask  con- 
taining 63  gallons.  Having  filled  up  the  cask  with  water,  he  draws  the 
same  quantity  as  before  and  then  finds  that  only  28  gallons  of  pure  wine 
remain  in  the  cask.     How  many  gallons  did  he  draw  each  time  ? 

16.  A  man  travels  50  miles  by  the  train  A,  and  then  after  a  wait  of  5 
minutes  returns  by  the  train  B,  which  runs  5  miles  an  hour  faster  than 
the  train  A.  The  entire  journey  occupies  2i  hours.  What  are  the  rates 
of  the  two  trains  ? 

17.  A  pedestrian  walked  6  miles  in  a  certain  interval  of  time.  Had 
the  time  been  1/2  hour  less,  the  rate  would  have  been  2  miles  per  hour 
greater.     Required  the  time  and  rate. 

18.  A  pedestrian  walked  12  miles  at  a  certain  rate  and  then  6  miles 
farther  at  a  rate  1/2  mile  per  hour  greater.  Had  he  walked  the  entire 
distance  at  the  greater  rate,  his  time  would  have  been  20  minutes  less. 
How  long  did  it  take  him  to  walk  the  18  miles  ? 

19.  From  the  point  of  intersection  of  two  straight  roads  which  cross 
at  right  angles,  two  men,  A  and  B',  set  out  simultaneously,  A  on  the  one 
road  at  the  rate  of  3  miles  per  hour,  B  on  the  other  at  the  rate  of  4  miles 
per  hour.     After  how  many  hours  will  they  be  30  miles  apart  ? 

20.  If  A  and  B  walk  on  the  roads  just  described,  but  at  the  rates  of 
2  and  3  miles  per  hour  respectively,  and  A  starts  2  hours  before  B,  how 
long  after  B  starts  will  they  be  10  miles  apart  ? 

21.  If  from  a  height  of  a  feet  a  body  be  thrown  vertically  upward  with 
an  initial  velocity  of  b  feet  per  second,  its  height  at  the  end  of  t  seconds 
is  given  by  the  formula  h  =  a  +  bt  —  16 1^.  The  corresponding  formula 
when  the  body  is  thrown  vertically  downward  is  ^  =  a  —  6^  —  16 1^. 

(1)  If  a  body  be  thrown  vertically  upward  from  the  ground  with  an 
initial  velocity  of  32  feet  per  second,  when  will  it  be  at  a  height  of  7  feet? 
of  16  feet  ?     Will  it  ever  reach  a  height  of  17  feet  ? 

(2)  A  body  is  thrown  from  a  height  of  64  feet  vertically  downward 
with  an  initial  velocity  of  48  feet  per  second.  When  will  it  reach  the 
height  of  36  feet  ? 

(3)  If  a  body  be  dropped  from  a  height  of  36  feet,  when  will  it  reach 
the  ground  ? 


304  A   COLLEGE   ALGEBRA 

XIV.     A   DISCUSSION    OF   THE    QUADRATIC 
EQUATION.     MAXIMA   AND    MINIMA 

635         Character  of  the  roots.     The  discriminant.     Let  a  and  y8  denote 

the  roots  of  ax"  +  ^x  +  c  =  0,  so  that,  §  631, 


h  4-  V/y-  ^A,ic       ^       -b  -  Vft^  -  4  ac 


2  a  2  a 

The  radicand  i^  —  4  ac  is  called  the  dlscrimbiant  of 
ax'^  -\-hx  -\-  c  —  Q. 
When  the  coefficients  a,  b,  c  are  real,  the  character  of  the  roots 
a,  p  is  indicated  by  the  sig7i  of  the  discriminant.     Thus  : 

1.  When  b^  —  4  ac  is  jjosifive,  the  roots  are  real  aiid  distinct. 

2.  When  b^  —  4  ac  is  0,  the  roots  are  real  and  equal. 

3.  When  b'^  —  4  ac  is  negative,  the  roots  are  conjugate  imagi- 
naries. 

It  should  also  be  observed  that 

1.  When  i^  —  4  ac  =  0,  then  ax"^  +  &a;  +  c  is  a  perfect  square. 

2.  When  a  is  positive  and  c  is  negative,  the  roots  are  always 
real,  since  b-  —  Aac  is  then  positive. 

3.  If  a,  b,  c  are  rational,  the  roots  are  rational  when,  and 
only  when,  i^  —  4  ac  is  a  perfect  square. 

Example  1.     Show  that  the  roots  of  x^  -  Gx  +  10  =  0  are  imaginary. 
They  are  imaginary  since  6'^  —  4  ac  =  (—  6)-  —  4  ■  1  ■  10  =  —  4. 

Example  2.     For  what  value  of  ??i  are  the  roots  of  mx-  +  3  x  +  2  =  0 
equal  ? 

We  must  have  S^  -  4  •  m  ■  2  =  0,  that  is,  ?«,  =  9/8. 

Example  3.     If  possible,  factor  y^  +  X7j  -  2x^  +  Ux  +  y  -  12. 

Arranging  the  polynomial  according  to  powers  of  y  and  equating  it  to 

0,  we  have  y^  +  (x  +  l)y  -  (2x2  -  11  x  +  12)  =  0. 

^  ,   .                             _  (X  +  1)  ±  V<)x2-42x  +  49 
Solvmg,  y  =  — '^ ■, 

that  is,  y  =  X  -  i,  or  y  -  -  2  X  +  S. 


A    DISCUSSION   OF    QUADRATIC    EQUATIONS     305 

Hence,  §  681,  2/2  +  xy  -  2  x2  +  11  x  +  ?/  -  12  =  (y  -  x  +  4)  (?/  +  2  x  -  3). 
Observe  that  the  factorization  is  possible  only  because  the  radicand 
9  x2  —  42  X  +  49  is  a  perfect  square. 

Relations  between  roots  and  coefficients.     If  a  and  /3  denote     636 
the  roots  of  ax^  -\-  bx  +  c  —  0,  we  have,  §  631, 

ax'^  -{-  bx  -\-  c  =  a[x  —  a)  (x  —  f3). 

Dividing  both  meniljers  of  this  identity  by  a  and  carrying 
out  the  multiplication  iu  the  second  member,  we  have 

b  c 

x^  +  -  X  +  -  =  x-  -  (a  +  B)x  +  aB. 

Since  this  is  an  identity,  the  coefficients  of  like  powers  of  x 
in  its  two  members  are  equal,  §  264,  that  is, 

a  -^  /3  =  —  b  I  a  and  a^  =  c  ]  a. 

This  may  also  be  proved  by  adding  and  multiplying  the 
values  of  a  and  ^  given  in  §  631.  Therefore,  since  a,  ^  are 
the  roots  oi  x'^  -^  bx  I  a  -^  c  j'  a  —  ^,  we  have  the  theorem : 

In  any  quadratic  of  the  form  x^  +  px  +  q  =  0  the  coefficient 
of  X  with  its  sign  changed  is  equal  to  the  sum  of  the  roots,  and 
the  constant  term  is  equal  to  the  product  of  the  roots. 

Thus,  in  the  quadratic  Gx^  +  x  =  2,  that  is,  x^  +  x/6  -  1/3  =  0,  the 
sum  of  the  roots  is  —1/6,  and  their  product  is  —  1  /3. 

Example  1.     Solve  Ox^  -  lOx  +  1  =  0. 

Obviously  one  of  the  roots  is  1,  for  9  —  10  +  1  =  0.  Therefore,  since 
the  product  of  the  roots  is  1/9,  the  other  root  is  1/9  -^  1  or  1/9. 

Example  2.  Find  the  equation  whose  roots  are  three  times  those  of 
3x2  +  8x  +  5  =  0. 

Let  a  and  j3  denote  the  roots  of  3  x^  +  8  x  +  5  =  0. 

Then  a  + /3  =  -  8/3  and  a/3  =  5/3. 

Hence  the  required  equation  is 
x2_  (3a  +  3/3)x  +  3rt-3i3  =  x2-  3(a  +  /3)x  +  9a/3  =  x2  +  8x  +  15  =  0. 

Symmetric  functions  of  the  roots.    The  expressions  a  -\-  (3  and     637 
afi  are  symmetric  functions  of  the  roots  a,  jS,  §  540.     All  other 
rational  symmetric  functions  of  a  and  0  can  be  expressed 


306  A    COLLEGE    ALGEBRA 

rationally  in  terms  of  these  two  functions,  a  +  ft  and  aft,  and 
therefore  rationally  in  terms  of  the  coefficients  of  the  equation. 

For  every  such  function  can  be  reduced  to  the  form  of  an  integral  sym- 
metric function  or  to  that  of  a  quotient  of  two  such  functions.  If  an 
integral  symmetric  function  contains  a  term  of  the  type  Zca''/3i' + ',  it  must 
also  contain  the  term  kaP  +  i^p,  §  5427lmtiUierefore  kaP^P {ai  +  /3'?).  But 
cxP^p  =  (cr/S)?,  and  it  may  readily  be  showl^■fe^jisuccessive  applications  of 
the  binomial  theorem  that  ai  +  /S"?  can  be  expresSed>4n^ terms  of  powers 
of  a  +  ^  and  a/3. 

Thus,  since  {a  4-/3)2=^2  +  2  n-/3  +  /32,  we  have  a^  +  p^  =  (a  +  /3)2  -  2^/3. 

Similarly  we  find  a^  +  /33  =  (a  +  /3)3  -  3  a/3  (a  +  /3). 

Example.  The  roots  of  x^  +  p,x  +  g  =  0  being  a,  j3,  express  1  /  a  +  1  /  /S 
and  a3/3  +  a/33  jn  terms  oi  p  and  q. 

We  have  l/a  +  1/^  =  (a  +  /3)/a/3=  -p/q, 
and  a3|3  +  a/33  =  ap(a^  +  ^2)  =a/3[(a  +  /3)2-2a^]=  g  (p2-2g). 

638  Infinite  roots.  Suppose  that,  instead  of  being  constants,  the 
coefficients  of  ax"^  -\-  bx  +  c  =  0  are  variables.  We  can  then 
show  that  if  a  approaches  0  as  limit,  one  of  the  roots  will 
approach  oo  ;  and  if  both  a  and  b  (but  not  c)  approach  0,  both 
roots  will  approach  x. 

For  the  formulas  for  the  roots  are 


-  b  +  V62  -  4  ac       ^       -b  -  Vb-  -  4  ac 


2  a 


Multiply  both  terms  of  the  fraction  a  by  —  6  —  V6-  —  4  ac  and  both 

terms  of  the  fraction  /3  by  -  6  +  V^"-  -  4  ac.  We  obtain 

2c  „  2c 

a  =  — 


5  +  Vb^  -4ac  b  -  V 62  _  4  ac 


By  §§  203,  205,       if  a  =  0,  then   V52  -  4  ac  =  b. 
Therefore  if  a  =  0,  then  a  ==  —  c/b  and  /3  =  00, 

and  if  a  =  0  and  6  =  0,  then  a  ==  00  and  /3  =  00. 

It  is  customary  to  state  these  conclusions  as  follows,  §  519: 
One  root  of  ax^  -f-  bx  +  c  =  0  becomes  injinite  when  a  van- 
ishes, and  both  roots  become  infinite  when  a  and  b  {Imt  not  c) 
vanish  simultaneously. 


A   DISCUSSION   OF   QUADRATIC    EQUATIONS     307 

Maxima  and  minima.     Let  ?/  be  a  function  of  x^  §  278.     It     639 
may  happen  that,  as  x  increases,  y  will  increase  to  a  certain 
value,  m,  and  then  begin  to  decrease,  or  that  y  will  decrease 
to  a  certain  value,  m',  and  then  begin  to  increase.     We  then 
call  m  a  maximum  value  of  y  and  m'  a  minimum  value. 

Thus,  y  =  {x  —  1)2  —  4  has  a  minimum  value  when  x  =  1,  this  value 
being  —  4.  For  if  x  start  from  a  value  less  than  1  and  increase,  (x  -  1)2 
will  first  decrease  to  0  and  then  increase. 

Similarly  y  =  4  —  (x  —  1)^  has  a  maximum  value,  4,  when  x  =  1. 

Every  quadratic  trinomial  ax^  -^hx  -\-  c  with  real  coefficients     640 
has  either  a  maximum  or  a  minimum  value,  which  may  be 
found  as  in  the  following  examples. 

Example  1.    Find  the  maximum  or  minimum  value  of  ?/  =  a;'-  +  6  x  —  7. 
By  completing  the  square,  x'-  +  6x  -  7  =  (x  +  3)2  —  10. 
Hence  when  x  =  —  3,  ?/  has  a  minimum  value,  namely  —  16. 

Example  2.  Divide  a  given  line  segment  into  two  parts  whose  rectangle 
shall  have  the  greatest  possible  area. 

Let  2  a  denote  the  length  of  the  given  segment,  x  and  2  a  —  x  the  lengths 
of  the  parts,  y  the  area  of  their  rectangle. 

Then  y  =  x  (2  a  -  x)  =  2  ax  -  x2  =  a2  -  (a  -  x)2. 

Hence  y  has  a  maximum  value  when  x  =  a,  that  is,  when  the  given  seg- 
ment is  bisected  and  the  rectangle  is  a  square  whose  area  is  a'. 

The  maximum  and  minimum  values  of  quadratic  trinomials     641 
and  of  certain  more  complex  functions  may  also  be  found  by 
the  following  method. 

Example.  Find  the  maximum  and  minimum  values,  if  any,  of 
2/  =  (4x2-2)/(4x-3). 

Clearing  of  fractions  and  solving  for  x,  we  have 


±  Vy2  _  3  y  ^-  2      2/ ±  V(y  -  1)  (2/ -  2) 


2  2 

By  hypothesis,  x  is  restricted  to  real  values.  Hence  y  can  only  take 
values  for  which  the  radicand  {y  —  1)  {y  —  2)  is  positive  (or  0),  that  is,  the 
value  1  and  lesser  values  and  the  value  2  and  greater  values. 

It  follows  from  this  that  1  is  a  maximum  and  2  a  minimum  value 
of  y. 


308 


A    COLLEGE    ALGEBRA 


For  observe  that  as  y  increases  to  1,  the  two  values  of  x,  namely 
2/  +  2)/2  and  (y  +  V?/2  _  3  ^  +  2)  /  2,  respectively  increase 
Hence,  conversely,  as  x  increases  through  1/2, 


(2/-V2/2 

and  decrease  to  1/2 

y  first  increases  to  1  and  then  decreases 


642         Variation  of  a  quadratic  trinomial.     Given  y  =  ax^  -\-hx  -{■  c, 
where  a  is  positive.     By  completing  the  square,  we  obtain 

Hence  3/  has  a  minimum  value  when  x  =—  b /2 a,  this  mini- 
mum value  being  (4ac  —  b^)/4:a.  (T 

As  X  increases  from  —  00  to  +  00,  ?/  will  first  decrease  from 
+  00  to  (i  ac  —  b^)  /  A  a  and  then  increase  to  +  00. 
Thus,  let  2/  =  x2  _  2  X  -  3  =  (a;  -  1)2  -  4. 

As  X  increases  from  —  00  to  +  <x>,  y  first 
decreases  from  00  to  —  4  and  then  increases 
from  —  4  to  00. 

Moreover  ?/  =  0  when  x^  _  2  x  —  3  =  0, 
that  is,  when  x  =  —  1  or  3. 

Until  X  reaches  the  value  —  1,  y  is  posi- 
tive ;  it  then  remains  negative  until  x  =  3, 
when  it  again  becomes  positive. 
When 

x= 3,-2,  -1,     0,      1,      2,3,4,   5,-.. 

we  have 

?/=•••   12,      5,      0,-3, -4, -3,0,5,12,. ••. 
We  may  obtain  the  graph  of  ?/  =  x^  —  2 x  —  3 
by  plotting  these  pairs  of  values  and  passing 
a  curve  through  them,  as  in  §  389. 

'(■\'_A\ 

^  '    }  Observe  that  to  the  zero  values  of  y  there 

correspond  the  points  where  the  graph  cuts  the  x-axis,  and  that  to  the 
minimum  value  of  y  there  corresponds  the  lowermost  point  of  the  graph, 
which  is  also  a  turning  point  of  this  curve. 


EXERCISE  XLin 

1.  For  what  valuesof  7?i  are  the  roots  of  (?«+2)x2- 2  >?ix  +  l=0  equal? 

2.  What  are  the  roots  of  {m'^  +  m)  x'-  +  3  mx  -2  =  0  when  m  =  -  1  ? 
when  ?w  =  0  ? 


EQUATIONS    SOLVABLE    BY   QUADRATICS        309 

3.  If  possible,  factor  Sx-  +  5xy  -  2y-  -  5x  +  'ty  -  2. 

4.  For  what  values  of  m  can  x-  -  y'^  +  mx  -{-  5y  —  6he  factored ? 

5.  The  roots  of  x'^  +  px  +  q  =  0  being  a  and  /3,  express  {a  -  /3)2, 
a*  +  /3*,  and  a/^  +  /3/a  in  terms  of  p  and  q. 

6.  The  roots  of2x2-3x  +  4  =  0  being  a  and  /3,  find  the  values  of 
a/p^  +  p/a^  and  a^^  +  ap^ 

7.  The  roots  of  a;-  +  x  +  2  =  0  being  rr  and  /3,  find  the  equations 
whose  roots  are  -  a,  -  )3 ;  1/a,  1/^;  2  a,  2/3;  a  +  1,  /3  +  1. 

8.  Find  the  maximum  and  minimum  values  of  the  following. 
1.    a;2_8x  +  3.  2.    2x2 -X +  4.  3.    l  +  4x-x2. 

4.   x/(x2  +  l).  5.    l/x  +  l/(l-x).  6.    (x  +  1)/ (2x2-1). 

9.  Find  the  greatest  rectangle  that  can  be  inscribed  in  a  given  circle ; 
also  the  rectangle  of  greatest  perimeter. 

10.  A  man  who  is  in  a  boat  2  miles  from  the  nearest  point  on  the 
shore  wishes  to  reach  as  quickly  as  he  can  a  point  on  the  shore  distant 
6  miles  from  that  nearest  point.  If  he  can  row  4  miles  an  hour  and 
walk  5  miles  an  hour,  toward  what  point  should  he  row  ? 

11.  "What  height  will  a  body  reach  if  thrown  vertically  upward  from 
the  ground  with  an  initial  velocity  of  48  feet  per  second,  and  when  will 
it  reach  this  height?     See  p.  303,  Ex.  21. 


XV.    EQUATIONS  OF  HIGHER  DEGREE  WHICH 
CAN  BE  SOLVED  BY  MEANS  OF  QUADRATICS 

Equations  which  can  be  factored.  Given  an  integral  equation  643 
in  the  form  A  =  0.  If  we  can  resolve  A  into  factors  of  the 
first  or  second  degrees,  we  can  find  all  the  roots  of  ^1  =  0  by 
equating  the  several  factors  of  A  to  zero  and  solving  the  result- 
ing equations.  For  if  ^  =  BC  •■■,  then  ^1  =  0  is  equivalent 
to  B=zO,  C  =  0,  ■•■,  jointly,  §  341. 

Example  1.     Solve  x*  +  x2  +  1  =  0. 

By  §  436,  X*  +  x2  +  1  =  (x2  +  X  +  1)  (x2  -  x  +  1). 


310  A    COLLEGE   ALGEBRA 

Hence  x*  +  x^  +  1  =  0  is  equivalent  to  the  two  equations 

x2  +  X  +  1  =  0  and  x2  -  x  +  1  =  0. 

„  ,  .       ^               .                  -1  ±  iVs-      1  ±  j.  Vs 
Solving  these  equations,      x  =  or 

Example  2.     Solve  x*  -  x^  -  5  x^  -  7  x  +  12  =  0. 
Factoring  by  the  method  of  §  451,  we  find  that 

x*  -  x3  -  5x2  -  7  X  +  12  =  (X  -  1)  (X  -  3)  (x2  +  3x  +  4). 
Hence  x*  —  x''  —  5  x^  —  7  x  +  12  =  0  is  equivalent  to  the  three  equations 
X  -  1  =  0,  X  -  3  =  0,  and  x2  +  3  X  +  4  =  0, 
whose  roots  are  1,  3,  and  (-  3  ±  i  V7)/2. 

Example  3.     Solve  the  following  equations. 

1.    6x3  -  11x2  + 8x -2  =  0.  2.    X* -5z3  +  x2  +  llx  +  4  =  0. 

644  Equations  of  the  type  au^  +  bu  +  c  =  0,  where  u  denotes  some 
function  of  x.  If  the  roots  of  au'^  +  bu  -{-  e  =  0  when  solved 
for  u  are  a  and  fi,  this  equation  is  equivalent  to  the  two 
equations  m  —  a  and  u  —  /?,  for,  §  631, 

au^  +  bu  -\-  c  =  a(u  —  a)  (u  —  (3). 
Hence  to  solve  au^  +  bu  +  c  —  0  for  x,  we  have  only  to  solve 
the  two  equations  u  =  a  and  u  =  (3  for  x. 

Example  1.     Solve  3  x<  +  10  x2  -  8  =  0. 

Solving  for  i2,  x^  =  2/3  or  -  4. 

Hence  x  =  ±Vt5/3  or  ±2i. 

Example  2.     Solve  x^  +  3  -  10x~^  =  0. 

Multiplying  by  x',     x^  +  3  x^  -  10  =  0. 

Solving  for  x',  x^  =  2  or  —  5. 

Hence  x  =  ±2V2or  ±5iV5. 

Example  3.     Solve  (x^  +  3  x  +  4)  (x2  +  3  x  +  5)  =  6. 
"We  may  reduce  this  equation  to  the  form 

(x2  +  3x)2  +  9  (x2  +  3x)  +  14  =  0. 
Solving  for  x2  +  3x,  we  obtain  the  two  equations 

x2  +  3  X  =  -  2,  and  x2  +  3  x  =  -  7, 
whose  roots  are         —  1,  —  2,  and  (—  3  ±  i  Vl9)/2. 


EQUATIONS    SOLVABLE    BY    QUADRATICS        311 

Example  4.     Solve  (x  +  1)  (x  +  2)  (x  +  3)  (x  +  4)  =  120. 

By  multiplying  together  the  first  and  fourth  factors,  and  the  second 
and  third,  we  reduce  the  equation  to  the  form 

(x2  +  5  X  +  4)  (X-  +  5  X  +  6)  =  120, 
which  may  be  solved  in  the  same  way  as  the  equation  in  Ex.  3. 

Example  5.     Solve  x^  +  10  x^  +  31  x^  +  30  x  +  5  =  0. 

By  completing  the  square  of  the  first  two  terms,  we  obtain 

{x2  +  5x)2  +  6  (x2  +  5x)  +  5  =  0. 
Solving  for  x^  +  5  x,  we  obtain  the  two  equations 

x2  +  5  X  =  —  5  and  x^  +  5  x  =  —  1, 
whose  roots  are  (-  5  ±  V6)/2  and  {-  5  ±  V2T)/2. 

Example  6.     Solve  8 ?i+l^  +  3  ^inl  _  n  =  0. 
x2  -  1  x2  +  2  X 

Observing  that  the  second  fraction  is  the  reciprocal  of  the  first,  we  mul- 
tiply both  members  of  the  equation  by  the  first  fraction,  thus  obtaining 


0. 


Solving  for  (x^  +  2x)  /(x^  —  1),  we  obtain  the  two  equations 

x2  +  2x      ,        ,x2  +  2x      3 

=  1  and =  - , 

x2  -  1  x2  -  1        8 

v7hose  roots  are  —1/2  and  —3,   —1/5. 

All  the  values  of  x  thus  found  are  roots  of  the  given  equation  since 

they  cause  none  of  its  denominators  to  vanish. 

Example  7.     Solve  the  following  equations. 

1.    3x*- 29x2 +  18  =  0.  2.   x''-6x3  +  8x2  +  3x  =  2. 

3.  (X  -  a)  (X  +  2  a)  (x  -  3  a)  (x  +  4  a)  =  24  a*. 

4.  (4x2  +  2x)/(x2  +  6)  +  (x2  +  G)/(2x2  +  x)  _  3  =  0. 

Reciprocal  equations.  These  are  equations  which  remain  645 
unchanged  when  we  replace  x  hy  1/x  and.  clear  of  fractions. 
If  we  arrange  the  terms  of  such  an  equation  in  descending 
powers  of  x,  the  first  and  last  coefficients  will  be  the  same, 
also  the  second  and  next  to  last,  and  so  on ;  or  each  of  these 
pairs  of  coefficients  will  have  the  same  absolute  values  but 
contrary  signs. 


312  A   COLLEGE    ALGEBRA 

Thus,  2x*  + 3x3 +  4x2  +  3x4-2  =  0 

and  x°-2x*  +  4x3-4x2  +  2x-l  =  0 

are  reciprocal  equations. 

Eeciprocal  equations  of  the  fourth  degree  may  be  reduced 
to  the  quadratic  form  and  solved  as  follows. 

Example  1.     Solve  2x*  -  Sx^  +  4x2  -  3x  +  2  =  0. 
Grouping  the  terms  which  have  like  coeflBcients  and  dividing  by  x2,  we 
reduce  the  given  equation  to  the  form 


<-'+^)-K'^+i)+^=''- 


X2. 

Since  x"^  -\-\  /x^  =  (x  +  1  /x)2  -  2,  we  may  reduce  this  equation  to  the 
torn,  J 


("=  +  i)-K^  +  i)  =  ''- 


Solving  for  x  +  1/x,  we  obtain  the  two  equations 

X  +  -  =  0  and  x  +  -  =  -, 
X  X      2 

whose  roots  are  i,   —  i,  and  (3±iv7)/4. 

Every  reciprocal  equation  of  odd  degree  has  the  root  1  or 
—  1 ;  and  if  the  corresponding  factor  a;  —  1  or  x  4-  i  be  sepa- 
rated, the  "depressed"  equation  will  also  be  reciprocal.  Hence 
reciprocal  equations  of  the  third  and  Jifth  degrees  can  be  solved 
by  aid  of  quadratics. 

Example  2.     Solve  2x5-3x2-3x  +  2  =  0. 

Grouping  terms,  2  (x"  +  1)  -  3  (x2  +  x)  =  0. 

Since  both  terms  of  this  equation  are  divisible  by  x  +  1,  it  is  equivalent 
to  the  two  equations 

X  +  1  =  0  and  2-x2  -  5  x  +  2  =  0, 
whose  roots  are  —1,  and  2,  1/2. 

Example  3.     Solve  x^  -  5  x*  +  9  x'  -  9  x2  +  5  x  -  1  =  0. 

Grouping  terms,  (x^  —  1)  —  5  x  (x^  —  1)  +  9  x2  (x  —  1)  =  0. 

Dividing  by  x  —  1,  we  find  that  this  equation  is  equivalent  to 

X  -  1  =  0  and  x*  -  4  x^  +  5  x2  -  4  x  +  1  =  0, 

whose  roots  are  1,  and  (1  ±  i  V3)/2,  (3  ±  V5)/2. 


EQUATIONS   SOLVABLE    BY    QUADRATICS        313 

Example  4.     Solve  the  following  equations. 

1.    x3  -  2X-2  +  2x  -  1  =  0.  2.   X*  -  4x3  +  5z2  _  4a;  +  1  =  q. 

3.    x5  +  x*  +  x^  +  x2  +  X  +  1  =  0. 

Binomial  equations.     This  name  is  given  to  equations  of  the     646 
form  x"  4-  a  =  0.     They  can   be  solved   by  methods  already 
given  when  cc"  +  a  can  be  resolved  into  factors  of  the  first  or 
second  degrees. 

Example  1.     Solve  x^  —  1  =  0. 

Since  x^  -  1  =  (x  -  1)  (x^  +  x  +  1),  the  equation  x^  -  1  =  0  is  equivalent 
to  the  two  equations 

X  -  1  =  0  and  x-  +  x  +  1  =  0. 

Solving,  x  =  1  or  (-  1  ±  i  V3)/2. 

Example  2.     Solve  x^  -  32  =  0. 

From  x5  —  32  =  0,  by  setting  x  =  V32  y  =  2  y,  we  obtain  ?/5  —  1  =  0. 

By  §§  438,  043,  t/^  —  1  =  0  is  equivalent  to  the  two  equations 
2/  -  1  =  0  and  2/4  +  2/3  +  2/-2  +  y  +  1  =  0. 

Solving,  2/  =  1,  (-  1  -£  V5  +  i  Vio  ±  2  V5)/4, 

or  (-  1  ±  V5  -  z  ViO  ±  2  V5)/4. 

Hence      x  =  22/  =  2,   (_l±V5  +  £  ViO  ±  2  V5)  /2, 
or  (-1  ±V5-i  ViO±  2  V5)/2. 

By  the  method  here  employed  every  binomial  equation  x"  ±  a  =  0  can 
be  reduced  to  the  reciprocal  form  y"  ±  1  =  0. 

Example  3.     Solve  the  following  equations. 

1.    x3  +  8  =  0.  2.    X*  +  1  =  0.  3.    x6  +  1  =  0. 

These  examples  illustrate  the  theorem  :  Every  number  has     647 
n  nth  roots.     Thus,  a  cube  root  of  1  is  any  number  which  satis- 
fies the  equation  x^  =  1\  and  in  Ex.  1,  we  found  three  such 
numbers,  namely  1,  (—  1  +  t  V3)/2,  and  (—  1  —  i  V3)/2, 

Irrational  equations.  If  asked  to  solve  an  irrational  equation,  648 
we  ordinarily  begin  by  rationalizing  it,  §  609.  But,  as  will  be 
illustrated  below,  certain  equations  admit  of  a  simpler  treat- 
ment than  this.  Whatever  method  is  used,  care  must  be  taken 
to  test  the  values  obtained  for  the  unknown  letter  before 
accepting  them  as  roots  of  the  given  equation. 


314  A    COLLEGE    ALGEBRA 


Example  I.     Solve  V2x  -3  -  Vox  -  6  +  V8~ 


Transposing,  v  2  x  —  3  +  v  3  x  —  5  =  v  6  x  —  6. 

Squaring  and  simplifying,  V(2  x  —  3)  (3  x  —  5)  =  1. 

Squaring  and  simplifying,  Gx^  —  19x  +  14  =  0. 

Solving,  X  =  2  or  7/6. 

Testing  these  values  of  x  in  the  given  equatinn,  we  find  that  2  is  a  root 
but  that  7/6  is  not  a  root. 

We  may  also  rationalize  the  given  equation  by  the  method  of  §  603. 
"We  thus  discover  that  —  4  (6x'^  —  19x  +  14)  is  identically  equal  to 


(V2x-3-V5x-6  +  V;^x-  5)(V2x  -3+V5x-6  -V3x-5). 

( V2x'^  +  V5X-6  +  V3x-5)(V2x-3  -  Vsx  -6  -  V3x-5). 
Tliere  are  but  two  values  of  x  for  which  the  product  6x2  —  19 x  +  14 
can  vanish.     The  first  factor  on  the   right  vanishes  for  one  of  these 
values,  2;  the  second,  for  the  other,  7/6.     Hence  there  is  no  value  of  x 
for  which  the  third  or  fourth  factor  can  vanisli. 


Example  2.     Solve  V4x  +  3  +  Vl2x  +  1  =  V24 x  +  10. 

Some  equations  which  involve  a  single  radical  can  be  reduced 
to  the  form  of  a  quadratic  with  respect  to  this  radical.  We 
then  begin  by  solving  for  tlie  radical. 


Example  3.     Solve  2  x2  -  6  x  -  5  Vx^  -  3  x  -  1  -5  =  0. 
Observing  that  the  x  terms  outside  the  radical  are  twice  those  under 
the  radical,  we  are  led  to  write  the  equation  in  the  form 

2  (x2  -  3  X  -  1)  -  5  Vx--!  -  3  X  -  1  -3  =  0. 
Solving  for  v'x2  —  3x  —  1,  we  obtain  the  two  equations 
Vx2-3x-l  =3,  and   Vx2  -  3x  -  1  =  -  1/2. 

The  second  of  these  equations  must  be  rejected  since  according  to  the 
convention  made  in  §  579  a  radical  of  the  form  ^/a  cannot  have  a  negative 
Talue. 

Squaring  the  first  equation,  we  obtain 

x2-3x-l  =  9, 
whose  roots  are  5  and  —  2. 

Testing  5  and  —  2  in  the  given  ecjuation,  we  find  that  both  of  them 
are  roots. 

Example  4.     Solve  2  x2  -  14  x  -  3  Vx2  -  7  x  +  10  +  18  =  0. 


EQUATIONS    SOLVABLE    BY    QUADRATICS        315 

Sometimes  an  equation  may  be  reduced  to  a  form  in  which, 
both  members  are  perfect  squares  or  one  member  is  a  perfect 
square  and  the  other  is  a  constant. 


Example  5.     Solve  Ax"  +  x  +  2x  VSxM-x  =  9. 
We  may  write  this  equation  in  the  form 

3  x2  +  X  +  2  X  V3  X-  +  X  +  x2  =  9. 
The  first  member  is  a  perfect  square,  and  extracting  the  square  root  of 
both  members,  we  obtain  the  two  equations 


Va  x2  +  X  +  X  =  3,  and   V3  x-  +  x  +  x  =  -  3. 
Solving  these  equations,  x  =  1,   —  9/2,  or  (5  ±  V97)  /4. 

Testing  these  results,  we  find  that  only  1  and  —  9/2  are  roots  of  the 
given  equation. 

Sometimes  all  the  terms,  when  properly  grouped,  have  a 
common  irrational  factor. 

Example  6.     Solve  Vx^—  7  ax  +  10  a-  —  Vx"-  +  ax  —  6  a^  =  x  —  2  a. 


Here  the  first  two  terms  and  also  x  —  2  a  have  the  factor  Vx  —  2  a. 
Separating  this  factor,  we  find  that  the  equation  is  equivalent  to 

Vx  —  2a  =  0  and  Vx  —  5a  —  Vx  +  3a  =  Vx  —  2 a. 
Solving  these  equations,         x  =  2a,  —  10 a /3,  or  6a. 

Testing,  we  find  that  only  2a  and  —10 a/3  are  roots  of  the  given 
equation. 


Example  7.     Solve  V3  x'^  -  5  x  -  12  -  V2  x^  -  11  x  +  15  =  x  -  3. 

If  one  or  more  of  the  terms  of  the  equation  are  fractions 
with  irrational  denominators,  it  is  often  best  to  rationalize 
these  denominators  at  the  outset. 

Example  8.     Solve  ( Vx  +  Vx  -3)  /  ( Vr  -  Vx  -  3)  =  2  x  -  5. 
Rationalizing  the  denominator  in  the  first  member  and  simplifying, 

we  have  

Vx2  -  3  X  =  2  X  -  6. 
Solving,  X  =  3  or  4. 

Testing,  we  find  that  both  3  and  4  are  roots  of  the  given  equation. 

Example  9.     Solve  ( Vx  -  1  -  Vx  +  1)  /  ( Vx  -  1  +  Vx  +  1)  =  x  -  3. 


316 


A   COLLEGE    ALGEBRA 


EXERCISE  XLIV 

Solve  the  following  equations. 

1.   4a;* -17x2 +  18  =  0.  2.    3x^-4x^  =  7. 

3.   (x2  -  4)  (x2  -  9)  =  7 x2.  4.   (2x2 -x-3)(3x2  +  x -2)2=0. 

5.   x<  +  x5  +  x2  +  3  X  -  6  =  0.  6.    X*  -  2  x3  +  x2  +  2  X  -  2  =  0. 

7.  (3x2-2x +  I)(3x2-2x- 7)  +  12  =  0. 

8.  x4  -  12  x3  +  33  x2  +  18  X  -  28  =  0. 

9.  4x*  +  4x3  -  x2  -  X  -  2  =  0. 

10.  X*  -  2  x3  +  2  x2  -  2  X  +  1  =  0. 

11.  x<  +  x3  +  2  X2  +  X  +  1  =  0. 

12.  x5  -  11 X*  +  36  x3  -  36  x2  +  11 X  -  1  =  0. 

13.  x5- 243  =  0.  14.    (2x-l)s  =  l. 

15.   (1  +  x)3  =  (1  -  x)3.  16.   (X  -  2)*  -  81  =  0. 

17.  (a  +  x)3  +  (6  +  x)3  =  (a  +  6  +  2  x)^. 

18.  (a  -  x)*  -  (6  -  x)4  =  (a  -  &)  (a  +  6  -  2  x). 
1 


^g     x2  +  3x  +  l       g4x2  +  6x 


4x2 +  6x 
„       1 


x2  +  3x  +  l 


-2 


21.  3x2 

22.  4x2 


2x-5V3x2-2x  +  3  +  9  =  0. 

2  X  -  1  =  V2  x2  -  X.         23.    V-r-^ 


31.    V5x2-6x  +  l  -  V5x2  +  9x-2 


SIMULTANEOUS   QUADRATIC   EQUATIONS        817 

32.    ^^^±^  +  3:.0.  33.    ^  +  4^2^  =  2. 

V2X-1  -  V3x 

34.  (X  +  a)5  +  (X  +  b}^  +  (X  +  c)'  =  0. 

35.  X  (X  -  1)  (x  -  2)  (X  -  3)  =  C  •  5  •  4  .  3. 

36.  (x  +  a)2  +  4  (X  +  rt)  Vx  =:  a-  -  4  a  Vx. 


XVT.     SIMULTANEOUS    EQUATIONS    WHICH 

CAN    BE    SOLVED    BY    MEANS    OF 

QUADRATICS 

A  PAIR  OF  EQUATIONS  IN  X,  Y,  ONE  OF  THE  FIRST 
DEGREE,  THE  OTHER  OF  THE  SECOND 

A  pair  of  simultaneous  equations  of  the  form  649 

f(x,  y)  =  ax^  +  bxi/  +  ci/'  -{- dx  +  ei/  +f  =  0, 
(f)  (ic,  y)  =  a'x  +  ^>'//  +  c'  =  0 
may  be  solved  as  in  the  following  example. 

Example.     Solve  y^  -  x- +  2x  +  2y  +  4  =0,  (1) 

2x-y-7  =  0.  (2) 

From  (2),                                                    ?/  =  2  x  -  7.  (3) 

Substituting  (3)  in  (1),     3  x2  -  22  x  +  39  =  0.  (4) 

Solving  (4),                                               X  =  13/3  or  3.  (5) 

Substituting  in  (.3),                                    y  =  5/3  or  -  1.  (6) 
The  solutions  of  (1),  (2)  are  the  pairs  of  values 

x=  13/3,  i/  =  5/3;  x  =  3,  y  =  -l.  (7) 

For,   §§368,  371,   the  following  pairs   of   equations   are  equivalent: 

(1),   (2)  to  (4),   (2);  (4),   (2)  to  (5),   (2);  (5),   (2)  to  (7). 

We  may  indicate  the  solutions  (7)  thus:  13/3,  5/3;  3,  —1.     Care 

must  always  be  taken  to  group  corresponding  values  of  x  and  y. 

Ordinarily  such  a  pair  of  equations  will  have  tico  finite  solu-     650 
tions.      But  if  the  group  of  first-degree  terms  in  ^  {x,  y),  namely 


318  A   COLLEGE   ALGEBRA 

a'x  +  Vy,  is  a  factor  of  the  group  of  second-degree  terms  in 
fix,  7j),  namely  ax^  +  bxy  +  cy";  while  (^  (x,  y)  itself  is  not  a 
factor  of  f{x,  y),  there  will  be  only  one  finite  solution  or  no 
such  solution.  And  if  ^(x,  y)  is  a  factor  of /(x,  y),  there  will 
be  infinitely  many  solutions. 

Example  1.     Solve        ?/'-  -  x^  +  2  a;  +  2  y  +  4  =  0,  (1) 

y  -mx  =  0.  (2) 

Eliminating  y,        {rrfi  -  1)  x2  +  2  (m  +  1)  x  +  4  =  0.  (3) 

If  11  fi  -1^0,  (3)  has  two  finite  roots,  and  (1),  (2)  two  finite  solutions. 

But  \iy  -  mx  is  a  factor  of  2/2  _  ^2,  that  is,  if  in  =  ±  1,  then  wi2  -1=0 
and  (3)  does  not  have  two  finite  roots. 

Thus,  if  jn  =  \,  (3)  reduces  to  x  +  1  =  0,  which  has  but  one  finite  root, 
the  other  being  infinite,  §  638.  And  if  m  =  -  1,  (3)  reduces  to  4  =  0, 
which  has  no  finite  root,  both  roots  being  infinite,  §  638. 

Hence  if  (2)  has  the  form  ?/  -  x  =  0,  the  pair  (1),  (2)  has  but  one  finite 
solution,  the  other  being  infinite.  And  if  (2)  has  the  form  ?/  +  x  =  0,  the 
pair  (1),  (2)  has  no  finite  solution,  both  solutions  being  infinite. 

Example  2.     Solve  y-  ~  x-  +  2x  +  2y  =  Q,  (1) 

y  +  x  =  0.  (2) 

Eliminating  y,  x^  -  x^  +  2x  -  2x  =  0.  (3) 

But  (3)  is  an  identity  and  is  satisfied  by  every  value  of  x. 

Hence  every  pair  of  numbers  x  =  a,  y  =  -aisa  solution  of  (1),  (2). 

The  reason  for  this  result  is  that  y  +  xisa,  factor  of  y^  -  x^  +  2x  +  2y. 

651  When  A,  B,  C  are  integral  functions,  the  pair  of  equations 
AB  ==0,  C  =  0  is  equivalent  to  the  two  pairs  A  =  0,  C  =  0 
and  5  =  0,  C  =  0,  §  371.  This  principle  and  §  649  enable  us 
to  solve  two  integral  equations  f(x,  y)  =  0,  (f>  (x,  y)  =  0  when- 
ever f(x,  y)  can  be  resolved  into  factors  of  the  first  or  second 
degrees  and  <^  {x,  y)  into  factors  of  the  first  degree. 

Example.     Solve  x^  +  xy-  —  bx  =  0),  (1) 

{2x-y)(x  +  y-\)  =  0.  (2) 
This  pair  of  equations  is  equivalent  to  the  four  pairs 

X  =  0,  2  X  -  2/  =  0,  (3) 

X  =  0,  X  +  y  -  1  =  0,  (4) 

x2  +  y2_5^0,  2x- j/  =  0,  '      (5) 

x2  -I-  2/2  _  5  =  0,  X  -I-  2/  -  1  =  0.  (6) 


SIMULTANEOUS   QUADRATIC    EQUATIONS  i     319 

Solving  the  pairs  (3),  (4),  (5),  (6),  we  find  the  soUitions  of  (1),  (2)  to  be 
0,  0 ;  0,  1 ;  1,2;  -  1,  -  2 ;  2,  -  1 ;  -1,2. 

A  pair  of  integral  equations  in  x,  y  can  be  solved  by  means     652 
of  quadratics  only  when  it  has  one  of  the  forms  described  in 
§§  649,  651  or  when  an  equivalent  pair  which  has  one  of  these 
forms  can  be  derived  from  it. 

Tims,  the  pair  of  equations  of  the  second  degree,  y^  —  x  +  1  =  0,  y  =  x% 
cannot  be  solved  by  quadratics.  For  there  is  no  simpler  method  of 
solving  this  pair  than  to  eliminate  y,  which  gives  x*  -  x  +  1  =  0,  an  equa- 
tion of  the  fourth  degree  which  cannot  be  solved  by  quadratics. 

The  preceding  sections  illustrate  the  truth  of  the  following     653 
important  theorem  : 

A  pair  of  integral  equations  f  (x,  y)  =  0,  <^  (x,  y)  =  0,  whose 
degrees  are  m  and  n  respectively,  has  mn  solutions. 

Thus,  the  pair  x^  +  x?/^  -  5  x  =  0,  (2  x  -  ?/)  (x  +  ?/  -  1)  =  0  has  3  •  2, 
or  6,  solutions,  §651.     See  §  381  also. 

It  should  be  added,  however,  that  if  the  groups  of  terms  of  654 
highest  degree  in  f{x,  y)  and  <l>(x,  y),  but  not  f{x,  y)  and 
<^(.r,  //)  themselves,  have  a  common  factor,  there  are  less  than 
mn  finite  solutions.  Thus,  for  every  factor  of  the  first  degree 
which  is  common  to  the  groups  of  terms  of  highest  degree  in 
f{x,  y)  and  <f>  (x,  y)  there  is  at  least  one  infinite  solution ;  for 
every  such  factor  which  is  also  common  to  the  groups  of  terms 
of  next  highest  degree  there  are  at  least  two  infinite  solutions; 
and  so  on.  lif(x,  y)  and  ^(x,  y)  themselves  have  a  common 
factor,  there  are  infinitely  many  solutions. 

Thus,  the  pair  x^  -  xy2  +  xy  _  2/2  -  y  =  0  (1),  x2  -  yS  _  i  =  o  (2)  cannot 
have  more  than  three  finite  solutions ;  for  there  are  3  •  2,  or  6,  solutions 
all  told,  and  at  least  one  of  these  is  infinite  since  x  +  ?/  is  a  common 
factor  of  the  groups  of  terms  of  highest  degree  in  (1)  and  (2),  namely 
x^  —  xy2  and  x"^  —  ?/'-,  and  at  least  two  others  are  infinite  since  x  —  y  is  a 
common  factor  of  the  groups  of  highest  and  next  highest  degree  in  (1),  (2), 
namely  x^  —  xy-^  x2  —  ?/2  and  xy  —  2/2,  0  (x  —  ?/)• 


j^     r.x2  +  y2-8  =  0,  ^^     p2_.ry_2^/2  +  y^0, 


320  A    COLLEGE    ALGEBRA 

EXERCISE   XLV 

Solve  the  following  pairs  of  equations. 
^      r7x2_6xy  =  8,       ^      ixy^l,  ^      rx2  +  x  =  4y2, 

■l2x-3?/  =  5.  ■     L3x-5y  =  2.         '     [Sx  +  dy^l. 

^     rSx2-Sxy-y"--ix-Sy  +  3  =  0, 
L3x  -?/-8  =  0. 

rx2  +  5y2_8x-72/  =  0,  r2x2-x?/-3y  =  0, 

■tx  +  3?/  =  0.  ■\7x-6y-4  =  0. 

rx2  +  3x?/  +  2;/2_  1  =  0,  J2x  +  3y  =  37, 

lx  +  ?/  =  0.  ■    ll/x  +  l/y  =  14/45. 

g      rl/z/-3/x  =  l,        ■  ^^     rx2  +  xy  +  2  =  0, 

L7/XZ/- 1/2/2  =  12..  ■    i(3x  +  2/)(2x  +  7/-l)  =  ( 

rx2 
(x  + 1)2  =  (2/ -1)2.  \(x-22/){x  +  2/-3)  =  0. 

13.  Determine  m  so  that  the  two  solutions  of  the  pair  ?/2  +  4  x  +  4  =  0, 
y  —  mx  shall  be  equal. 

14.  Determine  m  and  c  so  that  both  solutions  of  the  pair 

ic2  4-  .ry  -  2  2/2  +  X  =  0,  y  =  mx  +  c 
shall  be  infinite. 

15.  By  the  method  of  §  650,  Ex.  2,  show  that  2x  —  2/  +  4  is  a  factor 
of  2x2  +  X2/  -  2/2  +  10  X  +  2/  +  12. 

16.  Show  that  the  pair  x?/  =  1,  X2/  +  x  +  2/  =  0  has  not  more  than  two 
finite  solutions,  and  that  the  pair  x'~y  +  xy  =  \,  x-y  +  y"^  =  2  has  not  more 
than  four  finite  solutions. 


PAIRS   OF   EQUATIONS   WHICH   CAN   BE   SOLVED   BY 
FACTORIZATION,   ADDITION,   OR  SUBTRACTION 

655  When  both  equations  are  linear  with  respect  to  some  pair  of 
functions  of  x  and  y.  We  begin  by  solving  the  equations  for 
this  pair  of  functions  by  tlie  methods  of  §§  374-376. 

Example  1.     Solve  2x2  -  3^/2  =  — 58,  (1) 

3x2  +  2/-     =111.  (2) 


SIMULTANEOUS    QUADRATIC    EQUATIONS        321 

Solving  for  x^,  2/2,  we  obtain  x"-  =  25,  7/2  =  36, 
whence,  x  =  ±  5,  y  =  ±G. 

By  §§  367-372,  the  pair  (1),  (2)  is  equivalent  to  the  four  pairs  x  =  5, 
y  =  6;  x  =  —  5,  ?/  =  6;  x  =  5,  ?/=  —  6:  x  =  —  5,  ?/  =  —  6. 

Hence  the  solutions  of  (1),  (2)  are  5,  6  ;   -  5,  6  ;  5,  -  6 ;   -  5,  -  6. 


Example  2.     Solve  the  following  pairs. 

J ax2  +  by"  =  a,  jSx^  -l/y^  =  2, 

16x2  -ay'-  =  b.  "    1 5x2  +  3/^2  =  120. 


When  one  of  the  equations  can  be  factored.     This  is  always     656 
possible  when  the  equation  in  question  has  the  form 

ax^  -\-  bxy  +  cy"^  —  0, 
and,  in  general,  when  it  is  reducible  to  the  form 

air  -\-  bu.  -{-  c  =  0, 
where  u  denotes  a  function  of  x,  y. 

Example  1.     Solve  x2  +  y2  4.  x  -  11  ?/  -  2  =  0,  (1) 

x2- 5X2/ +  02/2  =  0.  (2) 
Factoring  (2)  by  solving  for  x  in  terms  of  j/, 

x  =  2y,  (3) 

or  '  x  —  Zy.  (4) 

Solving  (1),  (3)  and  (1),  (4),  we  obtain  all  the  solutions  of  (1),  (2), 
namely  4,  2;   -2/5,  -  1/5;  3,  1  ;   -3/5,  -  1/5. 

Example  2.     Solve  2x2  +  4x?/ +  2y2  _|_  3^  +  3  y  -  2  =  0,  (1) 

3  x2  -  32  (/2  +  5  =  0.  (2) 

We  may  write  (1)  thus :       2  (x  +  ?/)2  +  3  (x  +  y)  -  2  =  0. 

Solving,  x  +  2/=:]/2,  (3) 

or  X  +  2/  =  -  2.  (4) 

Solving  (2),  (3)  and  (2),  (4),  we  obtain  all  the  solutions  of  (1),  (2), 
namely  1,  -  1  /2  ;  3/29,  23/58  ;   -  3,  1 ;  -  41  /29,  -  17/29. 

Example  3.     Solve  the  following  pairs. 

^     rx2  +  x2/-6  =  o,  ^^    rx  +  .^x-.^^^ 

U2_2X2  =  1. 


322  A   COLLEGE   ALGEBRA 

657  When  the  given  equations  may  be  combined  by  addition  or  sub- 
traction so  as  to  yield  an  equation  which  can  be  factored.  This  is 
always  possible  when  both  the  given  equations  are  of  the 
form  ax^  +  i>^y  +  cy"^  =  d. 

Example  1.     Solve        6  x^  -  x?/  -  2  ?/2  =  56,  (1) 

5  x2  -  xy  -  y2  =  49.  (2) 

We  combine  (1)  and  (2)  so  as  to  eliminate  the  constant  terms. 

Multiply  (1)  by  7,    42  x2  _  7  xy  -  142/2  =  392.  (3) 

Multiply  (2)  by  8,      40  x2  -  8  xy  -  8  2/2  =  392.  (4) 

Subtract  (4)  from  (3),    2  x2  +  x?/  -  6  2/2  =  0.  (5) 

Solve  (5)  for  X,                                         x  =  3  2//2,  (6) 

or                                                                    X  =  —  2  2/.  (7) 

Solving  (2),  (6)  and  (2),  (7),  we  obtain  all  the  solutions  of  (1),  (2)* 

namely  ±3V36/10,  ±^^35/5;   ±2V2T/3,  tV^/3. 

And,  in  general,  we  obtain  an  equation  which  can  be  factored 
when  the  given  equations  are  of  the  second  degree,  and  can  be 
combined  by  addition  or  subtraction  so  as  to  eliminate  (1)  all 
terms  of  the  second  degree ;  (2)  all  terms  except  those  of  the 
second  degree ;  (3)  all  terms  which  involve  x  (or  y) ;  or  (4)  all 
terms  which  do  not  involve  x  (or  y). 

Example  2.     Solve  2  x2  +  4  X2/  -  2  x  -  ?/  +  2  =  0,  (1) 

3x2  +  Gx2/-x  +  32/  =  0.  (2) 

Here  we  can  eliminate  all  terms  of  the  second  degree  by  multiplying 
(1)  by  3,  and  (2)  by  2,  and  subtracting.     We  thus  obtain 

4x  +  92/-6  =  0.  (3) 

Solving  (2),  (3),  we  obtain  -  3.  2 ;  -  2,  14/0,  and  these  are  all  the 
finite  solutions  of  (1),  (2).     See  §  654. 

Example  3.     Solve      x2  -  3  x?/  +  2  2/2  +  4  x  +  3  ?/  -  1  =  0,  (1) 

2  x2  -  6  X2/  +  2/-  +  8  X  +  2  y  -  3  =  0.  (2) 

Here  all  the  terms  which  involve  x  can  be  eliminated  by  nmltiplying 
(1)  by  2  and  then  subtracting  (2).     We  thus  obtain 

32/2  + 4  2/ +  1  =0.  (3) 

Solving  (1),  (3),  wo  obtain  all  the  four  solutidns  of  (1),  (2),  namely 
1/3,  -  1/3;   -  1(5/3,  -  1/3;  (-  7  ±  V57)/2,  -  1. 


SIMULTANEOUS   QUADRATIC    EQUATIONS       323 

Consider  the  following  example  also. 

Example  4.     Solve  x-  +  2xy  +  2y^  +  ^x  =  0,  (1) 

xy  +  y^  +  2,y  +  \  =  0.  (2) 

Multiply  (2)  by  2  and  add  to  (1).     We  obtain 

(x  +  22/)2  +  3{x  +  22/)  +  2  =  0.  (3) 

Solving  (3),  x  +  22/  =  -l,  (4) 

or  a;  +  2  y  =  -  2.  (5) 

Solving  (2),  (4)  and  (2),  (5),  we  obtain  all  the  solutions  of  (1),  (2), 
namely  -  3  ±  2  V2,  1  T  ^2  ;   -  3  ±  ^5,  (1  =F  "^5)  /2. 
Example  5.     Solve  the  following  pairs  of  equations. 
^      r2x2  +  xy  +  52/=0,  rx2 +7/2  _  13=0, 

■     1x2  +  ?/2  +  10  ?/  =  0.  ■    t  xy  +  2/  -  X  =  -  1. 

"When  the  equation  obtained  by  eliminating  x  or  y  can  be  658 
factored.  From  any  pair  of  equations  of  the  second  degree 
we  can  eliminate  x  or  y  by  the  following  method.  The  result- 
ing equation  will  ordinarily  be  of  the  fourth  degree  and  not 
solvable  by  means  of  quadratics.  But  if  we  can  resolve  it 
into  factors  of  the  first  or  second  degrees,  we  can  solve  it  and 
so  obtain  the  solutions  of  the  given  pair. 

Example  1.     Solve      10  x2  +  5  ?/2  _  27  x  -  4  y  +  5  =  0,  (1) 

x2  +  ^2  _  3  X  _  y  =  0.  (2) 

First  eliminate  ?/2  by  multiplying  (2)  by  5  and  subtracting  the  result 
from  (1).     We  obtain 

5  x2  -  12  X  +  2/  +  5  =  0,  or  ?/  =  -  5  x2  +  12  X  -  5.  (3) 

Substituting  (3)  in  (2), 

5x* -24x3  + 40x2 -27x  + 6  =  0.  (4) 

Factoring,  by  §  45l, 

(X  -  1)  (X  -  2)  (5  x2  -  9  X  +  3)  =  0.  (5) 

Solving  (5),  §  643,  x  =  1,  2,  (9  ±  V21)/10.  (6) 

Substituting  (6)  in  (3),      ?/ =  2,  -  1,    (7  ±  3  \^)/10.  (7) 

The  pairs  of  corresponding  values  (6),  (7)  are  the  solutions  of  (1),  (2). 

Example  2.     Solve        x2  -  3  xy  +  2  ?/2  +  3  x  -  3  y  =  0, 
2  x2  +  a-y  -  y2  +  X  -  2  y  +  3  =  0. 


324  A   COLLEGE   ALGEBRA 

EXERCISE  XL VI 

Solve  the  following  pairs  of  equations. 

17x2-22/2  =  10.  ■    tl/?/2_4/j.2  =  8. 

^      ry^  +  xy  +  6  =  0,  ^     rx'^  +  y'^  -  3x  +  2y  -  39  =  0., 

'    \y2-y -2  =  0.  '    I3x2_i7a:y +  102/2  =  0. 

j'y2_x2-5  =  0, 

l4x2  + 4x2/ +  2/2 +  4x  + 2i 


x2+5x2/-2x  +  32/  +  l  =  0, 
3x2  +  15x2/-7x  +  82/  +  4  =  0, 
x2  -  15  X2/  -  3  2/2  +  2  X  +  9  2/  =  9 
5x2/  4-  2/-  -  3  2/  =  -  21. 
1x2 +  3x2/ -42/2  =  25,  fx{x  +  3y) 


r2x^  +  -6xy  -iy^  =  'Zb,  r:x(x  +  32/)  =  l 

1 15x2 +  24x2/ -31 2/2  =  200.  '    \x^-  ^y-  =  4. 


fx-  -  3 x?/  +  3 2/2  =  x22/2,  ja;2  +  X2/  +  2/-  =  38, 

1 7x2 -10x2/ +  4  2/2  =12x22/2.  '     1x2  -  x;/ +  ?/2  =  14. 

j'x2-x2/  +  y2  =  21(x-2/),  rx2  +  2/-8  =  0, 

1x2/-  =  20.  ■    1 2/2  +  15  X  -  40  =  0. 


10. 


PAIRS    WHICH   CAN   BE   SOLVED   BY   DIVISION 

659  In  solving  a  pair  of  equations  it  is  sometimes  advantageous 
to  combine  them  by  multiplication  or  division ;  but  care  must 
then  be  taken  not  to  introduce  extraneous  solutions  nor  to 
lose  actual  ones  (see  §§  362,  342). 

660  If  given  a  pair  of  the  form  AB  =.  CD  (1),  B  =  D  (2),  where 
A,  B,  C,  D  denote  integral  functions  of  x,  y,  we  may  replace  B 
by  D  in  (1),  thus  obtaining  the  pair  AD  =  CD,  B  =  D  which  is 
evidently  equivalent  to  the  two  pairs  .4  =  C,  B  =  D  and  D  =  0, 
J5  =  0.  We  may  obtain  the  pair  A  =  C,  B  =  D  by  dividing 
each  member  of  (1)  by  the  corresponding  member  of  (2);  but 
if  we  then  merely  solve  this  pair  A  =^  C,  B  —  D,  we  lose  some 


SIMULTANEOUS    QUADRATIC    EQUATIONS        325 

of  the  solutions  of  (1),  (2),  except,  of  course,  when  either  B  or 
Z)  is  a  constant,  so  that  the  pair  iJ  =  0,  Z>  =  0  has  no  solution. 

Example  1.     Solve           x*  -\-  x-y-  -\-  y*  =  2\,  (1) 

x2  +  X!/  +  7/2  =  7.  (2) 

Dividing  (1)  by  (2),            x"^  -  xy  +  y^  =  3.  (3) 

Solving  (2),  (3),  we  obtain  all  the  finite  solutions  of  (1),  (2),  namely 

2,  1 ;   -  2,  -  1 ;  1,  2  ;   -  1,  -  2.     See  §  654. 

Example  2.     Solve  x^  -  ?/3  ^  -  3  (x  +  1)  y,  (1) 

x2  +  xy  +  y-^  =  X  +  1.  (2) 

Dividing  (1)  by  (2),  x-y  =  -2,y.  (3) 

The  pair  (1),  (2)  has  the  same  finite  solutions  as  the  pair  (2),  (3)  and 

the  pair  x-  +  x?/  +  t/^  =  o  (4),  x  +  1  =  0  (5)  jointly.     And  the  solutions  of 

(2),  (3)  and  (4),  (5)  are  2,-1;   -  2/3,  1/3;   -  1,  (1  ±  i  V3)/2. 

Example  3.     Solve  (x  +  y)"  =  x,  (1) 

x2  -  t/2  =  _  e  y.  (2) 

Dividing  (1)  by  (2),        (x  +  2/)/(x  -  ?/)  =  -  x/6?y.  (3) 

Clearing  of  fractions,     x"^  +  bxy  +  Qy"^  =  0.  (4) 

The  pair  (2),  (4)  has  the  four  solutions  0,  0  ;  0,  0  ;  4,  -  2  ;  9/4,  -  3/4. 

The  process  by  which  (4)  was  derived  from  (1),  (2)  is  reversible  when 

a;,  y  have  the  values  4,-2  or  9/4,  —  3/4,  but  not  when  they  have  the 

values  0,  0.     Hence  this  reckoning  only  proves  that  4,-2  and  9/4, 

-3/4  are  solutions  of  (1),  (2),  §362. 

It  is  obvious  by  inspection  that  0,  0  is  a  solution  of  (1),  (2);  but  it 
should  be  counted  only  once  as  a  solution,  not  twice  as  in  the  case  of 
(2),  (4).  This  follows  from  the  fact  that  (1),  (2)  can  have  but  three  finite 
solutions,  §  654.  It  may  also  be  shown  thus  :  In  (1),  (2)  make  the  substi- 
tution y  =  tx  (5).  We  obtain  (1  +  ty^x"^  =  x  (U),  (1  -  t"^)^^  =  -  Qtx  (2'). 
And  (5),  (1'),  (2')  yield  x  —  0,  y  =  0  once  and  but  once. 

EXERCISE  XLVn 


^     rx3-y'  =  63,  2  rx  +  y  =  g8, 

rx*  +  x22/2  4-^4  =  931,  ^  r(x  +  2/)(x2-2i 

I  x2  +  x?/  +  ?/2  =  49.  ■  \(X-7J)  (x2  -  2  \ 

,      r(x  +  2/)2(x-?/)  =  3xy+6j/,  ^  j'x2-3xy  +  2y' 

Ix*-- j^2-a;  +  2.  "  lx2-2/2  =  -5y, 


326  A   COLLEGE    ALGEBRA 


SYMMETRIC   PAIRS   OF   EQUATIONS 

661         A  pair  of  equations  in  x,  y  is  said  to  be  symmetric  if  it 
remains  unchanged  when  x  and  y  are  interchanged. 


•   Thus,  the  following  pairs,  (a)  and  (6),  are  symmetric. 

IxV  +  xy  +  1  =  0.  ^'\y^  =  2y  +  Zx. 

Symmetric  pairs  are  of  two  types,  those  like  (a)  in  which 
the  individual  equations  remain  unchanged  when  x  and  y  are 
interchanged,  and  those  like  (b)  in  which  the  two  equations 
change  places  when  x  and  y  are  interchanged. 

662  Symmetric  pairs  of  the  first  type.  The  simplest  pair  of  sym- 
metric equations  is  x  -{-  y  =  a,  xy  =^h.  This  pair  may  be 
solved  as  in  §  649,  but  the  following  is  a  more  symmetric 
method. 

Example.     Solve  x  +  y  =  5,  (1) 

xy  =  6.  (2) 

Square  (1),  x'^  +  2xy  +  y^  =  25.  (3) 

Multiply  (2)  by  4,  4xy  =  24.  (4) 

Subtract  (4)  from  (3),       x"^  -  2xy  +  y'^  =  \.  (5). 

Hence  x  —  y  —  1,  (6) 

or  X  -  2/  =  -  1.  (7) 

From  (1),  (6),  x  =  3,  y  =  2 ;  and  from  (1),  (7),  x  =  2,  2^  =  3. 

663  If  given  a  more  complicated  pair  of  symmetric  equations, 
we  may  transform  each  equation  into  an  equation  in  x  +  y 
and  xy,  §  637,  and  then  solve  for  these  functions ;  or  in  the 
given  equations  we  may  set  x  =  ti  +  v,  y  =  u  —  v  and  then 
solve  for  u  and  v.  The  second  method  is  essentially  the  same 
as  solving  the  given  equations  for  x  -{-  y  and  x  —  y;  for,  since 
X  =  u  -{-  V,  y  =  u  —  t^,  we  have  «  =  (a;  -f-  y)  /2,  ^)  =(x  —  y)  /2. 

Example  1.     Solve    2x-  +  r^  xy  +  2y'^  +  x  +  y  +  \  =  0,  (1) 

x2  +  4x2/  +  2/2  +  12x  +  12?/  +  10  =  0.  (2) 

In  (1)  and  (2)  for  x^  +  y"^  substitute  (x  +  y)^  -  2xy. 


0, 

(3) 

0. 

(4) 

0. 

(5) 

4, 

(6) 

-2/3. 

0) 

-37, 

(8) 

-11/9. 

(9) 

SIMULTANEOUS   QUADRATIC    EQUATIONS        327 

Collecting  terms,  2{x  +  y)~  +  xy  +  (x  +  y)  +  1 

{X  +  y)^  +  2x!/  +  12  {x  +  y)  +  lO: 

Eliminating  xy,       S(x  +  y)^  -  10  (x  +  ?/)  -  8  : 

Solving,  X  +  y 

or  x  +  y  ■ 

Hence,  from  (3),  ,  xy  ■■ 

or  xy  ■■ 

And  solving  the  pairs  (6),  (8)  and  (7),  (9)  for  x,  y,  v?e  have 
X,  2/  =  2±Vil,  2t^^41;  (-1  ±2  V3)/3,  (-1  t2  V3)/3. 

Example  2.     Solve                      x*  +  y*  =  97,  (1) 

x  +  y  =  5.  (2) 
In  (1)  and  (2)  set             x  =  ii  +  v,  y  =  u  —  v. 

We  obtain                  (u  +  ?))*  +  (u  -  v)*  =  97,  (3) 

and                                                            2  M  =  5.  (4) 

Eliminating  u,          16  u*  +  600  v"-  -  151  =  0.  (5) 

Solving,                                                      v=:±l/2  or  ±iVl5T/2.  (6) 
Substituting  m  =  5  /  2  (4)  and  the  four  values  (6)  of  v  in  the  formulas 

X  =  u  +  V,  y  =  u  ~  V,  we  obtain 

X,  2/ =  2,  3;  3,  2;  (5  ±iVT51)/2,   (5  ^F  i  vl51)/2. 

Evidently  if  a:  =  cr,  ?/  =  /?  is  one  solution  of  a  symmetric 
pair,  X  =  1^,  y  =  ix  is  another  solution.  Unless  a  =  y8,  these 
two  solutions  are  different ;  but  xt/  and  x  +  i/  have  the  same 
values  for  x  =  a,  y  =  ^  as  for  x  =  (3,  y  —  a,  and  the  corre- 
sponding values  oi  X  —  y,  namely  a  —  jB  and  jS  —  a,  differ  only 
in  sign. 

Hence  the  values  of  xy  or  x  -{-  y  derived  from  a  symmetric 
pair  will  be  less  numerous  than  the  values  of  x  or  y,  that  is 
the  degree  of  the  equation  in  xy  or  x  +  y  derived  from  the 
pair  by  elimination,  as  in  Ex.  1,  will  be  less  than  the  degree 
of  an  equation  in  x  ov  y  similarly  derived  would  be.  As  for 
the  equation  in  x  —  y,  if  c  is  one  of  its  roots,  —  c  must  be 
another  root.  Hence  this  equation  will  involve  only  even 
powers  of  x  ~  y,  as  in  Ex.  2,  or  only  odd  powers  with  no 
constant  term. 


328  A    COLLEGE    ALGEBRA 

664  Note.  The  methods  just  given  are  applicable  to  pairs  of  equations 
which  are  symmetric  with  respect  to  x  and  -  y  or  some  other  pair  of 
functions  of  x  and  y.  Tlius,  x*  +  y*  =  o,  x-y  -h  may  be  written 
X*  +  {-yY  =  a,  x  +  {-y)  =  h. 

665  Symmetric  pairs  of  the  second  type.  Such  a  pair  may  some- 
times be  solved  as  follows. 

Example.     Solve  x'^Tx  +  Sy,  (1) 

2/3  =  7  2/  +  3x.  (2) 

Adding  (1)  and  (2),  x^  +  ?/3  =  ]0 (x  +  y).  (3) 

Subtracting  (2)  from  (1),      x/  -  2/3  =  4  (x  -  y).  (4) 

By  §  341,  (3)  is  equivalent  to  the  two  equations 

X  +  2/  =  0,  (5) 

and  X-  -  xy  ^  y-  =  10.  ^6, 

Similarly  (4)  is  equivalent  to  the  two  equations 

X  -  2/  =  0,  (7) 

and  X-  +  xy  -\-  y-  =  4.  (8) 

And  solving  (5),  (7)  ;  (5),  (8)  ;  ((1)^(7) ;  (6),  (8),  we  obtain  0,  0  ;  2^-  2  ; 

-2,    2;    ±VlO,    ±Vl0;    (1  ±  Vl3)/2,   (1  T  ^^)/2 ;    (-l±Vl3)/2, 

(_l^Vr3)/2. 

EXERCISE   XLVIII 
Solve  the  following  pairs  of  equations. 

rx  +  ?/  =  5,  rx2  +  2/2  =  200,      ^      rx2  +  2/2  =  293, 

\xy  +  m  =  0.  ■    tx  +  2/  =  12.  ■     1x2/ =  34. 

fx2  +  2/2  =  85,  rx'5  +  2/^  =  513,      ^      r  x^  +  J/'' =  468, 

■  Ix  -  2/  =  7.  Ix  +  y  =  9.  1x22/  +  x2/2  =  420 
(-27x3  +  04  2/3  =  65,            r  x"  +  2/*  =  82,        ^      rx5  +  2/^  =  32, 

■  l3xf42/  =  5.  ■     lx-2/  =  2.  '    tx  +  2/ =  2. 

rx  +  2/  =  l/2,  .a.y  +  ^4.y  +  19  =  0, 

■^^-     I  56('?  +  I)  +  113  =  0.  ■    1x22/  +  X2/2  +  20  =  0. 


x<  +  2/*  -  (-i;-  +  2/-)  =  72,  ^g      r  x22/  +  X2/2  =  30, 

2/2  +  2/2  =  10.  ■    ll/x+ 1/2/ =  3/10. 


j2      rx*  +  2/*-(.x2  +  2/2)  =  72,  ^g  (X'y  +  : 

■    1x2  +  j.22/2  +  ,y2  ^19.  •  \  1  /x  + 

rx2  +  3x2/  +  2/2  +  2x  +  2  2/=8,  ^^  r  x3  =  5  2/, 

l2x2  +  22/2  +  3x  +  32/  =  14.  "  1.2/3=  5x. 


SIMULTANEOUS   QUADRATIC   EQUATIONS        329 

SYSTEMS  INVOLVING  MORE  THAN  TWO  UNKNOWN  LETTERS 

A  system  of  three  equations  in  three  unknown  letters  can  be 
solved  by  means  of  quadratics  when  one  of  the  equations  is  of 
the  second  degree  and  the  other  two  of  the  first  degree ;  also 
when  it  is  possible  to  reduce  the  system  to  one  or  more  equiva- 
lent systems  each  consisting  of  one  equation  of  the  second 
degree  and  the  rest  of  the  first  degree.  The  like  is  true  of  a 
system  of  four  equations  in  four  unknown  letters,  and  so  on. 

If  A,  B,  C  are  integral  functions  of  degrees  m,  n^ji  in  x,  y,  z, 
and  no  two  of  them  have  a  common  factor,  the  equations 
A  =  0,  B  =  0,  C  =  0  will  have  vmp  solutions.  But  some  of 
these  solutions  may  be  infinite. 

Example  1.     Solve  z-  —  xy  —  1  =  Q,  (1) 

X  +  y  +  z  =  0,  (2) 

3  X  -  2  2/  +  2  z  +  2  =  0.  (3) 
Solving  (2),  (3)  for  x  and  y  in  terms  of  z, 

x  =  -(4z  +  2)/5,  (4) 

2/  =  (-z  +  2)/5.  (5) 
Substituting  in  (1)  and  simplifying, 

7  22  +  2  z  -  57  =  0.  (6) 

Solving  (6), 

Hence,  from  (4),  (5),  x, 

Example  2.     Solve  the  system 


Dividing  (2)  by  (1), 
Substituting  (4)  in  (3), 
From  (5),  (4),  (1)  we  obtain 


2  =-3  or  19/7. 
2/,  z  =  2,  1,  -3;   -y,    - 

1     19 

7'    7  ' 

X2/  =  6, 

(1) 

yz  =  12, 

(2) 

2X  =  8. 

(3) 

z/x  =  2  or  2  =  2x. 

(4) 

X-  =  4,  whence  x  =  ±  2. 

(5) 

2/,  z  =  2,  3,  4 ;   -  2,  -  3,  - 

-4. 

EXERCISE   XLIX 
Solve  the  following  systems  of  equations- 


rx  +  y  =  S, 
iy  +  z  =  2. 

-x(y  +  z)  =  12, 

2.    - 

2/{2  +  x)  =  6, 

1x2  _  2/2  ^  19. 

.2(X  +  2/)  =  10 

r{y  +  b){z  +  c)  =  a'^, 
«!  (2  +  c)(x  +  a)  =  62, 
l(x  +  a)(t/  +  6)=c«. 


330  A   COLLEGE    ALGEBRA 


EXERCISE  L 

Solve  the  following  systems  of  equations  by  any  of  the  methods  of  the 
present  chapter. 

\2z-3y  =  5.         "'    \x-y  =  l.  '   \xy  ^  {b'^  -  a'-) / 'i. 

( \      3 
(a       b        ^  ,  ,.,  =  1,  (x  +  y  =  a  +  b, 

[x2  +  2/2  =  0.  —  -  ~  =  12.  [x  +  b      y  +  a 

Ixy      2/2 

Lx      y       5  Lxy  t  4 


10.   a{x-\-y)  =  b(x-y)  =  xy.         11.   40x2/  =  21(x2-?/2)  =  210(x  +  ?/). 

■4x2-252/2  =  0,  rx2+3x2/-92/2  =  9, 

10 2/2 -3 2/ =  4.  ■    tx2- 13xy  +  2l2/2  =  -9. 


12     r4-^-25.'^  =  0,  ^3_     rx^- 

12x2-10  2/2 -3  2/ =  4.  1x2 

rx2-72/2-29  =  0,  rx/y  +  2//x  =  65/28, 

lx2-6xy  +  92/2-2x  +  62/  =  3.  "    1 2  (x2  +  2/2)  +  (x 


16. 


y)  =  34. 
rx22/  =  a^  j'x22/  +  X2/2  =  a,  rx  =  a(x2  +  2/2), 


^^      rx22/  +  X2/2  =  a,  ^g      re 

1x2/2  =  &.  ■     1x22/ -  X2/2  =  6.  ■     12/  -&{a;2  +  2/2), 

r(x  +  y)/(x-2/)  =  5/3,  ^^     r3(x3- y^)  ^  isx^,, 

l(2x  +  32/)(3x-22/)  =  110a2.         '     ix-y^l. 


19.     , 

.(2x  +  32/)(3x-2  2/)  =  110a2. 

2j      ,x*  +  2/*  =  aS  22      f21(x  +  2/)  =  10x2A 


rx*  + 

1x4-: 


y  =  a.  U  +  2/ +  a;2-f  2/2  =  68. 

23.   x2  +  2/2  =  X2/  =  X  +  2/.  24.   x2  -  X2/  +  2/^  =  3  a2  =  x2  - 

2^      rx2  +  x2/  +  2/-  =  21,  26      r4x2-32/2  =  12(x-2/), 

Ix  +  Vxy -f  2/ =  7.  '     1x2/ =  0. 

j'x2  +  2/2  =  x  + ?7-f20,  28      ra;-  +  4x-32/  =  0, 
lx2/  +  10  =  2(x  +  2/).  ■    li 


2/2  +  lOx  -  92/  =  0. 

X2/  —  x/y  =  a, 
y  =  2.  Ixy  -  y/x  =  l/a. 


r28(x6  +  2/5)  =  61{x3  +  y3),       ^^     rx2/ -  x/2/ =  a, 
■    lx  +  2/  =  2.  ■    I 


SIMULTANEOUS    QUADRATIC    EQUATIONS        331 

f{x  + 

\x  + 


3,_       (x  + 1)3 +  (2, -2)3  =  19,        ^^ 


y^  -  xy  -  yz  =  6, 

X  +  4  2/  +  z  =  14,  34. 

x-y  +  2Z-0. 


C{y  +  z){x-{-y  +  z)  =  10, 
35.    J  (z  +  x)  (X  +  ?/  +  z)  =  20,         36. 
l(x  +  ?/)(x  +  2/  +  z)  =  20. 


EXERCISE  LI 

1.  The  difference  of  the  cubes  of  two  numbers  is  218  and  the  cube  cf 
their  difference  is  8.     Find  the  numbers. 

2.  The  square  of  the  sum  of  two  numbers  less  their  product  is  63, 
and  the  difference  of  their  cubes  is  189.     What  are  they  ? 

3.  Tlie  sum  of  the  terms  of  a  certain  fraction  is  11,  and  the  product 
of  this  fraction  by  one  whose  numerator  and  denominator  exceed  its 
numerator  and  denominator  by  3  and  4  respectively  is  2/3.  Find  the 
fraction. 

4.  Separate  37  into  three  parts  whose  product  is  1440  and  such  that 
the  product  of  two  of  tliem  exceeds  three  times  the  third  by  12. 

5.  The  diagonal  of  a  rectangle  is  13  feet  long.  If  each  side  were  2 
feet  longer  than  it  is,  the  area  would  be  38  square  feet  greater  than  it  is. 
What  are  the  sides  ? 

6.  The  perimeter  of  a  right-angled  triangle  is  36  inches  long  and  the 
area  of  the  triangle  is  54  square  inches.     Find  the  lengths  of  the  sides. 

7.  The  hypotenuse  of  a  right-angled  triangle  is  longer  than  the  two 
perpendicular  sides  by  3  and  24  inches  respectively.  Find  the  sides  of 
the  triangle. 

8.  Find  the  dimensions  of  a  room  from  the  following  data  :  its  floor 
is  a  rectangle  whose  area  is  224  square  feet,  and  the  areas  of  two  of  its 
side  walls  are  126  and  144  square  feet  respectively. 

9.  A  rectangle  is  surrounded  by  a  border  whose  width  is  5  inches. 
The  area  of  the  rectangle  is  168  scjuare  inches,  that  of  the  border  360 
square  inches.     Find  the  length  and  breadth  of  the  rectangle. 


332  A    COLLEGE   ALGEBRA 

10.  In  buying  coal  A  gets  3  tons  more  for  -$135  than  B  does  and  pays 
$7  less  for  4  tons  than  B  pays  for  5,  Kequired  the  price  each  pays 
per  ton. 

11.  A  certain  principal  at  a  certain  rate  amounts  to  $1248  in  one  year 
at  simple  interest.  "Were  the  principal  $100  greater  and  the  rate  1|  times 
as  great,  the  amount  at  the  end  of  2  years  would  be  $1456.  What  is  the 
principal  and  what  is  the  rate  ? 

12.  A  man  leaves  $60,000  to  his  children  and  grandchildren,  seven  in 
all.  The  children  receive  ^  of  it,  .which  is  $2000  more  apiece  than  the 
grandchildren  get.  How  many  children  are  there  and  how  many  grand- 
children, and  what  does  each  receive  ? 

13.  At  his  usual  rate  a  man  can  row  15  miles  downstream  in  5  hours 
less  time  than  it  takes  him  to  return.  Could  he  double  his  rate,  his  time 
downstream  would  be  only  1  hour  less  than  his  time  upstream.  What 
is  his  usual  rate  in  dead  water  and  what  is  the  rate  of  the  current  ? 

14.  Three  men  A,  B,  C  together  can  do  a  piece  of  work  in  1  hour, 
20  minutes.  To  do  the  work  alone  it  would  take  C  twice  as  long  as  A 
and  2  hours  longer  than  B.  How  long  would  it  take  each  man  to  do  the 
work  alone  ? 

15.  Two  bodies  A  and  B  are  moving  at  constant  rates  and  in  the  same 
direction  around  the  circumference  of  a  circle  whose  length  is  20  feet. 
A  makes  one  circuit  in  2  seconds  less  time  than  B,  and  A  and  B  are 
together  once  every  minute.     What  are  their  rates  ? 

16.  On  two  straight  lines  which  meet  at  right  angles  at  0  the  points 
A  and  B  are  moving  toward  0  at  constant  rates.  A  is  now  28  inches 
from  0  and  B  9  inches  ;  2  seconds  hence  A  and  B  will  be  13  inches  apart, 
and  3  seconds  hence  they  will  be  5  inches  apart.  At  wliat  rates  are  A 
and  B  moving  ? 

17.  Three  men  A,  B,  and  C  set  out  at  the  same  time  to  walk  a  certain 
distance.  A  walks  4\  miles  an  hour  and  finishes  the  journey  2  hours 
before  B.  B  walks  1  mile  an  hour  faster  than  C  and  fiuislies  tlie  journey 
in  3  hours  less  time.     What  is  the  distance  ? 

18.  Two  couriers  A  and  B  start  sinmltaneously  from  P  and  Q  respec- 
tively and  travel  toward  each  otlier.  When  they  meet  A  has  traveled 
12  miles  farther  than  B.  After  their  meeting  A  continues  toward  Q  at 
the  same  rate  as  before,  arriving  in  4i;  hours.  Similarly  B  arrives  at  P 
in  7^  hours  after  the  meeting.     What  is  the  distance  from  P  to  Q  ? 


SIMULTANEOUS    QUADRATIC    EQUATIONS        333 


GRAPHS   OF  EQUATIONS   OF   THE   SECOND  DEGREE   IN  X,  Y 

Examples  of  such  graphs.     The  graph  of  any  given  equation     667 
of  the  second  degree  in  .r,  y  may  be  obtained  by  the  method 
illustrated  in  the  following  examples. 

Example  1.     Find  the  graph  of     y- —  \x.  (1) 

Solving  for  y,  2/  =  ±  2  Vx.  (2) 

From  (2)  it  follows  that  when  x  is  negative,  y  is  imaginary  ;  when  x  is 
0, 2/  is  0  ;  when  x  is  positive,  y  has  two  real  values  which  are  equal  numeri- 
cally but  of  opposite  signs.  Hence  the  graph  of  (1)  lies  entirely  to  the 
right  of  the  y-axis,  passes  through  the  origin,  and  is  symmetric  with 
respect  to  the  x-axis. 

When  x  =  0,   1/4,      1/2,        1,  2,  3,        4,..., 

we  have  ?/  =  0,     ±  1,    ±  ^2,    ±  2,    ±  2  V2,   ±  2  Vs,    ±4,  •  •  • . 

We  obtain  the  part  of  the  graph  given  in  the  figure  by  plotting  these 
solutions  (0,  0),  (^,  1),  (],  —  1),  •  •  •  and  passing  a  curve  through  the 
points  thus  found.  Com- 
pare §  389.  It  touches  the  ^ 
y-axis.  y/^/        ^-^9.6) 

This  curve  is  called  a 
parabola.  It  consists  of 
one  "in  finite  branch,  "here 
extending  indefinitely  to 
che  right. 

Example  2.  In  what 
pohits  is  the  graph  of 
2/-  =  4  X  (1)  met  by  the 
graphs  of  y  =  x  —  S  (2), 
?/=x  +  l(3),  y  =  x  +  3(4)? 

1.  The  solutions  of  (1), 
(2)  are  1,  -  2 ;  9,  6.     Hence,  §  386,  the  graphs  of  (1),  (2)  intersect  in  the 
points  (1,  —  2),  (9,  6),  as  is  .shown  in  the  preceding  figure. 

2.  The  solutions  of  (1),  (3)  are  equal,  namely  1,2;  1,  2.  Hence 
the  graph  of  (3)  meets  the  graph  of  (1)  in  two  coincident  points  at  (1,  2). 
Tliis  means  that  the  graph  of  (3)  touches  the  graph  of  (1)  at  (1,  2),  as  is 
indicated  in  the  figure. 

3.  The  solutions  of  (1),  (4)  are  imaginary,  namely  —  l±2V2i, 
2  ±  2  V2  i.     Hence,  as  the  figure  shows,  the  graphs  of  (1),  (4)  do  not  meet. 


334 


A   COLLEGE   ALGEBRA 


Example  3.     Find  the  graph  of 

2/2 -2x2/ +  2x2- 

Solving  for  y,  we  have      y  =  x  - 


1X  +  22/  +  1 


-l±V4x-x2.  (2) 

The  values  of  y  given  by  (2)  are  real  when  the  radicand  4  x  —  x2,  or 
x(4  —  x),  is  positive  (or  0),  that  is,  vfhen  x  lies  between  0  and  4  (or  is  0 
or  4).     Hence  the  graph  of  (1)  lies  between  the  lines  x  =  0  and  x  =  4. 

When  X  =  0  and  when  x  =  4  the  values  of  y  given  by  (2)  are  equal: 
namely  —  1,-1  when  x  =  0,  and  3,  3,  when  x  =  4.  Tliis  means  that  the 
graph  of  (1)  tmches  the  line  x  =  0  at 
the  point  (0,  —  1)  and  the  line  x  =  4 
at  the  point  (4,  3).  See  Ex.  2,  2.  The 
line  y  —  X  ~\  passes  through  these 
points  of  tangency,  for  when  4x  —  x- 
vanishes,  (2)  gives  the  same  values  for 
y  that  y  =  X  —  1  gives. 

For  each  value  of  x  between  0 

and  4  the  equation  (2)  gives  two  real 

_Y   and  distinct  values  of  y,  obtained  by 

increasing  and  diminishing  the  value 

of  X  —  1  by  that  of  V4  x  —  x2.     Hence 

for  each  of  these  values  of  x  there  are 

two  points  of  the  graph  of  (1).     They 

^  are  most  readily  obtained  by  drawing 

the  line  y  =  x  —  \  and  then  increasing  and  diminishing  its  ordinate  fur 

the  value  of  x  in  question  by  tiie  value  of  V4  x  —  x^. 

Thus,  when  x  =      0,  1,  2,  3,  4, 

we  have  for  the  line  ?/  =  — 1,  0,  1,  2,  3, 

and  for  the  graph  of  (1)     2/  =  -  1,    ±  V^,  1  ±  2,  2  ±  V3,  3. 

The  figure  shows  the  oval-shaped  curve  which  the  points  thus  found 
determine.     It  is  called  an  ellipse. 

By  solving  (1)  for  x  and  applying  the  method  of  §  641,  we  may  show 
that  the  highest  and  lowest  points  of  the  curve  are  (2  +  V2,  1  +  2  V2) 
and  (2  -  V2,  1  -  2  y.^2). 

Example  4.     Find  the  graph  of  t/'^  -  x-  +  2  x  +  2  ?/  +  4  =  0.  (1) 

Solving  for  y  and  factoring  the  radicand  in  the  result, 

2/  =  -  1  ±  V(x  +  1)  (X  -  3).  (2) 

The  radicand  (x  +  1)  (x  —  3)  vanishes  when  x  =  —  1  and  when  x  =  3, 

and  in  both  cases  (2)  gives  equal  values  of  y,  namely  —  1,  —  1.     This 


SIMULTANEOUS   QUADRATIC    EQUATIONS        335 


means  that  the  graph  of  (1)  touches  the  line  x  =  —  1  at  the  point  (—1,  —  1) 
and  the  line  x  =  3  at  the  point  (3,  —  1).  The  line  y  =  —  1  passes  through 
these  points  of  tangency. 

The  radicand  (x  +  1)  (x  —  3)  is  positive  when  and  only  when  x  <  —  1 
or  X  >  3.  For  every  such  value  of  x  the  equation  (2)  gives  two  real 
and  distinct  values  of  y  and  therefore  two  points  of  the  graph  of  (1). 
These  points  may  be  obtained  by  drawing  the  line  y  —  —  1  and  then 
increasing  and  diminishing  its  con- 
stant  ordinate  —  1  by  the  value  of 
V(x  +  l)(x-3). 

Hence,  as  is  indicated  in  the 
figure,  the  graph  of  (1)  consists  of 
two  infinite  branches,  the  one  touch- 
ing the  line  x  =  —  1  and  extending 
indefinitely  to  the  left,  the  other 
touching  the  line  x  =  3  and  extend- 
ing indefinitely  to  the  right. 

This  curve  is  called  an  hyperbola. 

There  are  two  straight  lines  called 
asyinptotes,  which  the  infinite 
branches  of  this  hyperbola  tend  to  touch,  and  which  they  are  said  to 
touch  at  infinity.  These  lines  are  the  graphs  of  the  equations  ?/  =  x  —  2 
and  y  =  —  X,  which  we  obtain  as  follows.     Compare  §  650,  Ex.  1. 

Eliminating  y  between  (1)  and  the  equation  y  =  nix  +  c,  (3) 

we  obtain     (??i2  _  i)  x2  +  2  {mc  +  m  +  1)  x  +  (c^  +  2  c  +  4)  =  0.  (4) 

Both  roots  of  (4)  are  infinite,  §  638, 
if  m-  —  1  =  0  and  mc  +  m  +  1  =  0, 

that  is,  if  m  =  1,  c  =  —  2,  or  if  ?n  =  —  1,  c  =  0. 

Hence  both  solutions  of  (1),  (3)  are  infinite  if  (3)  has  one  of  the  forms 
y  =  x-2     (3')        or        y  =  -  x.     (3") 

Therefore  the  graphs  of  (3')  and  (3")  each  meet  the  graph  of  (1)  in  tiuo 
injiniiely  distant  coincident  points. 

Example  5.     Find  the  graph  ot  y\- 4x1/  +  3x^  +  6x  -  2y  ^  0.       (1) 

Solving  for  y,  we  have    y  =  2x  +  1  ±  Vx2  -  2x  +  1,  (2) 

that  is,  ?/  =  3  X  or  ?/  =  X  +  2. 

Hence  the  graph  of  (1)  consists  of  the  pair  of  right  lines  y  =  5x  and 
y  =  x  +  2. 

Except  when  the  radicand  vanishes,  that  is,  when  x  —  1  =  0,  the  equa- 
tion (2)  gives  two  real  and  distinct  values  of  y.     But  when  x  —  1  =  0  it 


336 


A   COLLEGE   ALGEBRA 


' 

(-4,JA^'^ 

0 

1) 

\A 

J 

^X^_ 

::::::: 

^^-3) 

gives  two  equal  values  of  y,  namely  3,  3.  Hence  the  line  x  —  1  =  0 
meets  the  graph  of  (1)  in  two  coincident  points  at  (1,  3).  Of  course  this 
cannot  mean  that  the  line  x  -  1  =  0  touches  the  graph  of  (1)  at  (1,  3).  It 
means  that  the  points  coincide  in  which  the  line  x  —  1  =  0  meets  the  two 
lines,  y  =  Zx  and  y  =  x  +  2,  which 
together  constitute  the  graph  of  (1). 
Example  6.  Find  the  graphs,  and 
their  intersections,  of 

x2  +  if-  =  25,  (1> 

xV16  +  2/V9==2.  (2) 

The  graph  of  (1)  is  a  circle  whose 

center  is  at  the  origin,  0,  and  whose 

radius  is  5.     The  graph  of  (2)  is  an 

ellipse. 

These  curves  intersect  at  the  four 
-"  points   (4,  3),  (-4,  3),  (-4,   -3), 

(4,  —  3),  these  points  being  the  graphs  of  the  solutions  of  (1),  (2). 
Example  7.     Find  the  graphs,  and  their  intersections,  of 
xy  -  S  y  -  2  =  0, 
xy  +  2y  +  5  =  0. 
From  (1)  we  obtain 
Here  there  is  one  real  value  of  y 
for  each  real  value  of  x. 

When     x  =  -  M,   -1,        0,        1,        2,      2] 
■we  have      y  =  0,        -  i,   -  f ,   -  1,   -  2,   -  8,   ±  <»,     8,  2,    0. 

And  plotting  these  solutions,  we  obtain  an  hyperbola  whose  two  infinite 
branches  are  indicated  by  the  unbroken  curved  lines  in  the  figure,  and 
whose  asymptotes  are  y  =  0  (found 
as  in  Ex.  4),  and  x  —  3  =  0  (since, 
when  X  =  3,  then  y  =  oo). 

In  a  similar  manner,  we  find  the 
graph  of  (2)  to  be  the  hyperbola  indi- 
cated by  the  dotted  curve  and  having 
the  asymptotes  y  =  0  and  x  +  2  =  0. 
The  equations  (1),  (2)  have  the 
single  finite  solution  x  =  1,  ?/  =  —  1, 
the  remaining  tliree  solutions  being 
infinite,  §  654. 

The    hyperbolas   which  are    the 
graphs  of  (1),  (2)  meet  in  the  single  finite  point  (1, 


0) 
(2) 

^  =  2/(x-3).  (3) 

d  hence  one  point  of  the  graph, 


H, 


But  since  they 


SIMULTANEOUS    QUADRATIC    EQUATIONS        337 

have  the  common  asymptote  y  =  0,  they  are  regarded  as  having  two 
infinitely  distant  coincident  intersections  at  (oc,  0) ;  and  since  they  have 
the  parallel  asymptotes  a;  —  3  =  0  and  x  +  2  =r  0,  they  are  regarded  as 
having  one  infinitely  distant  intersection  at  (0,  co). 

General  discussion  of  such  graphs.     Generalizing  the  results     668 
obtained  in  the  preceding  exaui])les,  we  reach  the  following 
conclusions. 

Suppose  any  equation  of  the  second  degree  in  x,  y  given, 
with  real  coefficients,  as 

ax"  +  2  hxxj  +  hif  +  2  ryro^  +  2/y  +  c  =  0.  (1) 

If  h  is  not  0,  and  we  solve  for  ?/,  we  have 

hy  =  -{lix^f)±-jR,  (2) 

where       R  =  (A^  -  ab)  x' +  2  (hf  -  brj)  x  +  (f^  -  be). 

From  (2)  we  obtain  two  real  values  of  y  for  each  value  of  x 
for  which  the  radicand  R  is  positive.  Corresponding  to  these 
two  values  of  y  there  are  two  points  of  the  graph  which  may- 
be found  by  drawing  the  line 

by^-ihx+f) 

and  then  increasing  and  decreasing  its  ordinate  for  the  value  of 
X  in  question  by  the  value  of  -y/lt/b.     See  §  667,  Exs.  1,  3,  4. 

The  form  of  the  graph  depends  on  the  character  of  the 
factors  of  R. 

1.  When    (hf  -  byf  -  (A^  -  ah)  {f  -  be)  =  0. 

In  this  case  jR  is  a  perfect  square,  §  635,  and  the  first 
member  of  (1)  can  be  resolved  into  factors  of  the  first  degree, 
§  635,  Ex.  3,  If  these  factors  have  real  coefficients,  the  graph 
of  (1)  is  a  pair  of  right  lines.     See  §  667,  Ex.  5. 

2.  When    {hf  -bg^-  {h^  -  ab)  (f  -  be)  >  0. 

In  this  case,  unless  h^  —  ab  =  0,  the  radicand  R  can  be 
reduced  to  the  form  R  =  (h^  —  ab)  (x  —  a)  (x  —  /3)  (3)  where 
(I  and  (3  are  real  and  a  <  ^,  §  635. 


338  A   COLLEGE    ALGEBRA 

If  h^  —  ab  <  0,  the  product  (3)  is  positive  when  and  only 
when  X  lies  between  a  and  (3.  Hence  the  graph  of  (1)  will  be 
a  closed  curve  lying  between  the  lines  x  —  a  =  0  and  x  —  /3  =  0, 
which  it  touches.  It  is  therefore  an  ellijjse  (or  circle).  See 
§  667,  Ex.  3. 

If  7^2  _  (jj  >  0,  the  product  (3)  is  positive  when  and  only 
when  x<a  or  x  >  ^.  Hence  the  graph  will  consist  of  two 
infinite  branches,  the  one  touching  the  line  x  —  a  =  0  and 
extending  to  its  left,  the  other  touching  the  line  x  —  (3  =  0 
and  extending  to  its  right.  It  is  therefore  an  hyperbola.  See 
§  667,  Ex.  4. 

If  h^  -ab  =  0,  we  have  R  =  2  {hf  -  bg)  x+{p-  be),  where 
kf—bg4^  0,  and  this  is  positive  when  and  only  when  we  have 
x-> —  (f^  —  be) /2(hf— bg).  Hence  the  graph  will  consist 
of  one  infinite  branch  lying  entirely  to  one  side  of  the  line 
2  (hf—  bg)  X  +  {f'^  —  be)  =  0,  which  it  touches.  It  is  therefore 
^  2)arabola.     See  §  667,  Ex.  1. 

3.    When    (hf-bg)^-(h^-ab)(f^-be)<0. 

In  this  case  the  roots  of  R  =  0  are  conjugate  imaginaries, 
§  635,  and  if  we  call  them  X  +  fii  and  A  —  /xi,  we  can  reduce  R 
to  the  form  R  =  (h^  -  ab)l(x  -  X)^  +  fx^^  (4). 

If  h^  —  ab>0,  the  product  (4)  is  positive  for  all  values  of  x. 
Hence  the  graph  of  (1)  will  consist  of  two  infinite  branches 
which  lie  on  opposite  sides  of  the  line  bi/  —  —  (hx  +/)•  It  is 
therefore  an  hi/jwrbola.     Thus,  i/"^  —  x"  —  1. 

If  K^  —  ab<  0,  the  product  (4)  is  negative  for  all  values 
of  X.  Hence  the  graph  of  (1)  will  be  wholly  imaginary.  Thus, 
x^  +  y'  +  1  =  0. 

In  the  preceding  discussion  it  is  assumed  that  bi^O.  But 
if  ^  =  0,  while  a  ^  0,  and  we  solve  (1)  for  x  instead  of  y,  we 
arrive  at  similar  conclusions.  If  both  a  =  0  and  b  =  0,  the 
graph  of  (1)  is  an  hyperbola,  as  in  §  667,  Ex.  7,  or  a  pair  of 
straight  lines  of  which  one  is  parallel  to  the  cc-axis,  the  other 
to  the  y-axis. 


SIMULTANEOUS   QUADRATIC    EQUATIONS       339 

EXERCISE  LII 

Find  the  graphs  of  the  following  equations. 
1.    2/2  =  -  8  X.  2.    x2  +  2/2  =  9.  3.    (y  _  x)"^  =  z. 

4.    x2  +  2x2/  +  2y2  =  8.  5.    y2  -  4x?/ +  3x2  +  4x  =  0. 

6.    y'^  -2xy  +  1-0.  7 .    y'^  -  2xy  -  1  =  0. 

8.    2  x2  +  3  2/2  _  4  X  +  6  2/  =  0.  9.    2/^  -  x2  -  3  x  +  2/  -  2  =  0. 

10.  2  x2  +  4  X2/  +  4  2/2  +  X  +  4  2/  -  5  =  0. 

11.  4  x2  -  12  X2/  +  9  2/2  +  3  X  -  6  y  =  0. 

Find  the  graphs  of  the  following  pairs  of  equations  and  their  points  of 
intersection. 


12. 


15. 


16. 


rx2/  =  l,  ^3_     rx2- 2/2  =  1,  ^^     rx2  +  2/^ 

l3x-52/  =  2.  Ix2-x2/  +  x  =  0.  12/2=  2z 


j'2/2-X2/-2x2-2z-22/-2  = 
L2/2-X2/-2x2  +  2  =  0. 
•  (X  -  2  2/)  (X  +  2/)  +  X  -  3  2/  =  0, 
:2/)(x-2/)  +  2x-62/  =  0. 


r(x-2i 
l(x-2. 


17.  Find  the  graph  of  x2  +  2/2  —  6x  —  22/  +  l=0  and  its  points  of 
intersection  with  the  axes  of  x  and  y. 

18.  Show  that  the  graph  of  {x  —  y)-  -  2{x  +  y)  +  I  =  0  touches  the  x 
and  2/  axes. 

19.  Show  that  the  line  y  =  Sx  +  5  touches  the  graph  of  Wx"  +  y"  —  \G  =  0 
at  the  point  (-3/5,  16/5). 

20.  For  what  values  of  m  will  the  line  y  =  mx  +  3  touch  the  graph  of 
x2  +  2  y2  =  G  ? 

21.  For  what  values  of  c  will  the  line  7x  —  4y  +  c  —  0  touch  the 
graph  of  3  x2  —  2/2  +  X  =  0  ? 

22.  Show  that  the  lines  y  =  0  and  x  —  2  y  +  1  =  0  are  the  asymptotes 
of  the  graph  of  X2/  —  2  2/2  +  2/  +  6  =  0. 

23.  Find  the  asymptotes  of  the  graph  of  the  equation 

2  x2  +  3  X2/  -  2  y2  ^  a;  +  2  2/  +  2  =  0. 

24.  For  what  values  of  X  is  the  graph  of  x2  +  \xy  +  y'^  =  x  an  ellipse  ? 
a  parabola  ?   an  hyperbola  ? 


340  A   COLLEGE   ALGEBRA 


XVII.     INEQUALITIES 

669  Single  inequalities.  An  absolute  inequality  is  one  like 
^'  +  y'  +  1  >  0  which  holds  good  for  all  real  values  of  the 
letters  involved  ;  a  conditional  inequality  is  one  like  x  —  1  >  0 
which  does  not  hold  good  for  all  real  values  of  these  letters, 
but,  on  the  contrary,  imposes  a  restriction  upon  them. 

670  The  principles  which  control  the  reckoning  with  inequalities 
are  given  in  §  261.  From  these  principles  it  follows  that  the 
sign  >  or  <  connecting  the  two  members  of  an  inequality 
will  remain  unchanged  if  we  transpose  a  term,  with  its  sign 
changed,  from  one  member  to  the  other,  or  if  we  multiply  both 
members  by  the  fiduiwe  positive  number  ;  but  that  the  sign  >  will 
be  changed  to  <,  or  ince  versa,  if  we  multiply  both  members 
by  the  same  negative  number. 

Example  1.  Prove  that          a2  +  6^  >  2  ah. 

We  have  (a  -  6)2  >  0, 

that  is,  a2  -  2  a6  +  tfi  >  0, 

and  therefore  cfi  +  lf->2  ah. 

Example  2.  Prove  that  a"^  +  h'^  +  c^  >  ah  +  he  +  ca. 

We  have  cC^  +  h'^>2ah,  62  +  c2  >  2  be,  c2  +  a2  >  2  ca. 

Adding  the  corresponding  members  of  these  three  inequalities  and 
dividing  the  result  by  2,  we  have  a'^  -\-  b^  +  c'^>ah  +  be  +  ca. 

Example  3.     Solve  the  inequality  3x  +  5>x  +  11,  that  is,  find  what 
restriction  it  imposes  on  the  value  of  x. 

Transposing  terms,  2  x  >  6, 

whence  x  >  3. 

Example  4.     Solve  x2  -  2  x  -  3  <  0. 

Factoring,  (x  +  1)  (x  -  3)  <  0. 

To  satisfy  this  inequality  one  factor  must  be  positive  and  the  other 
negative.     Hence  we  nmst  have  x  >  —  1  and  <  3,  that  is,  —  1  <  x  <  3. 

671  Simultaneous  inequalities.  A  system  of  one  or  more  ine- 
qualities of  the  form 

ax  -{-  hy  +  c  >  0 


INEQUALITIES 


341 


may  be  solved  for  the  variables  x,  y  hj  a,  simple  graphical 
method  which  is  based  upon  the  following  consideration  : 

Draw  the  straight  line  which  is  the  graph  ot  ax  +  bi/  +  c  =  0, 
§  385.  Then  for  all  i)airs  of  values  of  x,  y  whose  graphs  lie 
on  one  side  of  this  line  we  shall  have  ax  +  by  -\-  €>  0,  and 
for  all  pairs  whose  graphs  lie  on  the  other  side  of  the  line  we 
shall  have  ax  -\-  by  -^  c  <.  0. 

Thus,  let  (xi,  yi)  be  a  point  on  the  graph  oi  y  —  (mx  +  c)  =  0  so  that 
yi  —  {mxi  +  c)  =  0.  Then,  if  1/2  <  Vi  so  that  the  point  (xi,  2/2)  lies  below 
the  line,  we  have  yo  —  (mxi  +  c)  <  0,  and  if  y^  >  2/1  so  that  the  point  (xi,  2/3) 
lies  above  the  line,  we  have  v/3  —  {mxi  +  c)  >  0. 

Example.     Solve  the  simultaneous  inequalities 

ki  =  x-2y  +  K0,  ^o  =  x  +  y-5<0,  k3  =  2x-y  -  1>0. 

Find  the  graphs  of  ki  =  0,  ko  =  0,  ks  =  0,  as  indicated  in  the  figure. 

The  inequality  A:2  <  0  is  satisfied  y 

by  those  pairs  of  values  of  x,  y  whose 
graphs  lie  on  the  side  of  the  line 
A;2  =  0  toward  the  origin ;  for  when 
X  =  0,  2/  =  0,  we  have  A;2  =  —  5,  that 
is  <  0.  It  may  be  shown  in  a  similar 
manner  that  the  inequalities  ki<0 
and  A;3>0  are  .satisfied  by  the  pairs 
of  values  of  x,  y  whose  graphs  lie  on 
the  sides  of  the  lines  ki  =  0,  k^  =  0 
remote  from  the  origin. 

Therefore   the   given   inequalities 
^1  <  0,  A:2  <  0,  ^"3  >  0  are  satisfied  by 
the  pairs  of  values  of  x,  y  whose  graphs  lie  within  the  triangle  formed 
by  the  three  lines. 

EXERCISE  LIII 

In  the  following  examples  the  letters  a,  b,  c  are  supposed  to  denota 
unequal  positive  numbers. 

1.  Trove  that  a/6  +  6/a> 2. 

2.  Prove  that  (a  +  b)  (a^  +  b"^)  >  (a2  +  62)2. 

3.  Prove  that  a^  +  6^  >  a'^b  +  ab". 

4.  Prove  that  a"b  +  b"a  +  W-c  +  d^b  +  c'^a  +  a^o  6  abc. 


342  A   COLLEGE   ALGEBRA 

5.  Prove  that  a^  +  63  +  c^  >  3  ahc. 

6.  Solve  X  +  7>3x/2  -  8. 

7.  Solve  2  x2  +  4  X  >  x-^  +  6  X  +  8. 

8.  Solve  (X  +  1)  (X  -  3)  (X  -  6)  >  0. 

9.  Solve  7/-x-2<0,  x-3<0,  y  +  l>Oby  the  graphical  method. 

10.  Also  ?/-x>0,  2/-2x<0. 

11.  Also  x  +  ?/  +  3>0,  ?/-2x-4<0,  y  +  2x  +  4>0. 

12.  Prove  that  x^  +  2  x  +  5  >  0  is  true  for  all  values  of  x. 

13.  Solve  x2  +  2/2  _  1  <  0,  2/2  _  4  x  <  0  by  a  graphical  method. 

XVIII.     INDETERMINATE    EQUATIONS    OF 
THE    FIRST    DEGREE 

672  Single  equations  in  two  variables.     Given  any  equation  of  the 

form 

ax  -\-oy  =  c 

where  a,  b,  c  denote  integers,  of  which  a  and  b  have  no  common 
factor.  We  seek  an  expression  for  all  pairs  of  integral  values 
of  X  and  y  which  will  satisfy  this  equation ;  also  such  of  these 
pairs  as  are  positive  as  well  as  integral. 

673  Theorem  1.  All  equations  ax  +  by  =  c  of  the  Idnd  just 
described  have  integral  solutions. 

For  since  a  and  h  are  prime  to  one  another,  by  the  method  explained 
in  §  491  we  can  find  two  integers  p  and  7,  positive  or  negative,  such  that 
ap  -\-hq  —  \  and  therefore  a  (pc)  +  h  (qc)  =  c,  and  this  proves  that  x  =  pc, 
y  —  qc  is  a,  solution  of  az  +  by  —  c. 

674  Theorem  2.  If  x  =  x,,,  y  =  y^  he  one  integral  solution  of 
such  an  equatio7i  ax  +  by  =  c,  all  of  its  integral  solutions  are 
given  by  the  formulas 

X  =  Xo  +  bt,  y  =  yo  -  at 

whe7i  all  possible  integral  values  are  assigned  to  tc 


INDETERxMiNATE    EQUATIONS  343 

First,  X  =  Xo  +  bt,  y  =  yo  —  at  is  always  a  solution  of  ax  +  by  —  c.     (1) 

For,  substituting  iu  (1),  a  (Xq  +  bt)  +  b(yo  —  at)  =  c, 
or,  simplifying,  axo  +  byo  =  c, 

which  is  true  since,  by  hypothesis,  x  =  Xo,  2/  =  2/o  is  a  solution  of  (1). 

Second,  every  integral  solution  of  (1)  is  given  hy  x  =  Xo  +  bt,  y  =  yo  —  at. 

For  let  X  =  Xi,  y  =  yi  denote  any  second  integral  solution. 

Then  axi  +  byi  —  c  and  axo  +  byo  =  c, 

whence,  subtracting,  b  (yi  —  yo)  =  —  a  (Xi  —  Xq).  (2) 

From  (2)  it  follows  that  &  is  a  factor  of  the  product  of  the  integers  a 
and  Xi  —  Xo.  Therefore,  since  b  is  prime  to  a,  it  must  be  exactly  con- 
tained in  Xi  —  Xo,  §  492,  1,  and  if  we  call  the  quotient  f,  we  have 

Xi  —  Xo  =  bt',  or  Xi  =  Xo  +  bt\  (3) 

And  substituting  (3)  in  (2)  and  simplifying,  we  also  have 

2/1  =  2/0-  at\  (4) 

From  §  §  673,  674  it  follows  that  every  equation  ax  -\-hy  =  c  675 
of  the  kind  just  described  has  infinitely  many  integral  solu- 
tions. When  a  and  h  have  contrary  signs  there  are  also 
infinitely  many  positive  solutions  ;  but  when  a,  and  b  have  the 
same  sign  there  is  but  a  limited  number  of  such  solutions  or 
no  such  solution. 

Thus,  one  solution  of  2  x  +  3  y  =  18  is  x  =  3,  ?/  =  4. 
Hence  the  general  solution  is  x  =  3  +  3  <,  i/  =  4  —  2  i. 
The  positive  solutions  correspond  to  <  =:  —  1,  0,  1,  2  and  are  x,  y  =  0,  6  ; 
3,  4  ;  6,  2  ;  9,  0. 

The  theorem  of  §  674  enables  one  to  write  down  the  general  676 
integral  solution  of  an  equation  of  the  kind  under  considera- 
tion as  soon  as  a  particular  solution  is  known.  A  particular 
solution  may  often  be  found  by  inspection.  Thus,  one  solution 
of  10  a;  +  3  2/  =  12  is  a;  =  0,  ?/  =  4.  A  particular  solution  may 
always  be  found  by  the  method  indicated  in  §  673 ;  also  by 
the  method  illustrated  in  the  following  example. 

Example.     Find  the  integral  solutions  of  7x  +  19 y  =  213.  (1) 

Solving  for  the  variable  with  the  smaller  coefficient,  here  x,  and  reducing, 
we  have 

x  =  ^li:=i^  =  30-2y  +  ^-^.  (2) 


344  A   COLLEGE   ALGEBRA 

Hence  if  x  is  to  be  an  integer  when  y  is  one,  (3  -  5?/)/ 7  must  be  an 
integer.     Call  this  integer  m,  so  that  (3  -  5  ?/)  /  7  =  u. 

Then  5  y  +  7  u  =  3.  (3) 

Treating  (3)  as  we  have  just  treated  (1),  we  have 

3-7M  3-2«  .,. 


2/  = 


5 

Set  (3  -  2  m)/ 5,  which  must  be  integral,  equal  to  v. 
Then  2  w  +  5  «  =  3.  (5) 

Treating  (5)  as  we  have  already  treated  (1)  and  (3),  we  have 

„  =  ^Jil-"==i-2.  +  l:^.  (6) 

2  2 

When  v-1  the  fractional  term  (l-r)/2  vanishes  and  u  has  the 
integral  value  —  1. 

Substituting  it  =  -  1  in  (4),  we  obtain  ?/  =  2. 
Substituting  y  =  2  in  (2),  we  obtain  x  =  25. 
Hence  the  general  integral  solution  of  (1)  is 

x  =  25  +  19«,  2/  =  2-7<. 
There  are  two  positive  solutions  corresponding  to  t  —  -\  and  t  =  0 
respectively,  namely  :   x,  y  =  6,  9  ;  25,  2. 

Observe  that  in  the  fractional  terms  of  (2),  (4),  (6)  the  numerical  values 
of  the  coefficients  of  y,  u,  v,  namely  5,  2,  1,  are  merely  the  successive 
remainders  occurring  in  the  process  of  finding  the  greatest  common 
divisor  of  the  given  coefficients  7  and  19.  We  finally  obtain  the  remainder, 
or  coefficient,  1,  because  7  and  19  have  no  common  factor.  The  like  will 
be  true  if  we  apply  the  method  to  any  equation  ax  +  by  =  c  in  which  a 
and  b  have  no  common  factor.  Hence  the  method  will  always  yield  a 
solution  of  such  an  equation. 

But  in  practice  it  is  seldom  necessary  to  complete  the  reckoning  above 
indicated.  Thus,  having  obtained  (4),  we  might  have  observed  that  m  =  -  1 
will  make  (3  -  2u)/5  integral,  which  would  have  at  once  given  us  y  =  2 
and  therefore  x  =  25,  by  (2). 

677  Observe  that  an  equation  ax  -{-by  =  c  with  integral  coeffi- 
cients of  which  a  and  b  have  a  common  factor,  as  d,  can  have 
no  integral  solution  unless  d  is  also  a  factor  of  c.  For  if  x  and 
y  were  integers,  d  would  be  a  factor  of  ax  +  by  and  therefore 
of  c.     Thus,  4  a;  +  6  y  =  7  has  no  integral  solution. 


INDETERiMINATE    EQUATIONS  345 

Simultaneous  equations.     The  following  example  will  illus-     678 
trate  a  method  for  tindiug  the  integral  solutions,  if  there  be 
any,  of  a  pair  of  simultaneous  equations  in  three  variables 
with  integral  coefficients. 

Example.     Find  the  integral  solutions  of 

Sx  +  Gy-2z  =  22,  (1) 

5x  +  8y  -(3Z  =  28.  (2) 

First  eliminate  z  and  simplify  the  resulting  equation. 

We  obtain  2  x  +  5  ?/  =  19.  (3) 

Next  find  the  general  solution  of  (3),  as  in  §  676. 

We  obtain  x  =  7  +  [,t,  y  =  1  -  2t.  (4) 

Next  substitute  (4)  in  (1)  and  simplify  the  result. 

We  obtain  2z-St  =  5.  (5) 

Next  find  the  general  solution  of  (5). 

We  obtain  2  =  1  -  3  m,  «  =  -  1  -  2  u,  (6) 

where  u  denotes  any  integer  whatsoever. 

Finally  substitute  i  =  —  1  —  2m  in  (4)  and  simplify. 

We  obtain       x  =  2  —  lOu,  y  — 3  +  4^u,  z  =  l  —  Su,  (7) 

which  is  the  general  solution  required. 

The  only  positive  solution  is  that  corresponding  to  m  =  0,  namely  x  =  2, 
2/ =  3,  2  =  1. 

Observe  that  the  given  equations  will  have  no  integral  solu- 
tion if  either  of  the  derived  equations  in  two  variables  has 
none,  §  677. 

We  proceed  in  a  similar  manner  if  given  three  equations  in 
four  variables,  and  so  on. 

Single  equations  in  more  than  two  variables.     The  following     679 
example  illustrates  a  method  of  obtaining  formulas  for  the 
integral   solutions    of    a    single    equation   in  more  than  two 
variables  with  integral  coefficients. 


Example.     Find  the  integral  solutions  of  5a;  +  Sy  +  19  2  =  50.         (1) 

Solving  for  X,  x  =  10  -  y  -  3  2  -  'l^jtJ^.  (2) 

5 

Set  {3y  +  iz)/5,  which  must  be  integral,  equal  to  u. 


346  A   COLLEGE   ALGEBRA 


Then  3  y  +  4  2  =  5  u.  (3) 

Solving  for  2/,  y  =  u-z+     "~^-  (4) 

Set  (2  u  —  2)/3,  which  must  be  integral,  equal  to  V. 

Then  z-2u  -  3v.  (5) 

Substituting  (5)  in  (4),    y  —  —  u  +  iv.  (6) 

Substituting  (5)  and  (6)  in  (2), 

x  =  10-6«  +  5u.  (7) 

The  formulas  (5),  (6),  (7),  in  which  u,  v  may  have  any  integral  values 
whatsoever,  constitute  the  general  solution  required. 

Substituting  u  =  2,  v  =  1  in  the  formulas  (5),  (6),  (7),  we  obtain  a 
positive  solution  of  (1),  namely  x  =  3,  y  =  2,  z  =  1. 

EXERCISE  LIV 

Find  the  general  integral  solutions  of  the  following ;  also  the  positive 
integral  solutions. 

1.  6x-172/  =  18.  2.    43x-12y  =  158. 

3.  16x  +  59y  =  l.  4.    72x  +  23  2/  =  845. 

5.  49x-27y  =  2S.  6.    47 x  -  97 y  =  50L 

^  r2x  +  52/-8z  =  27,  r5x  +  2y  =  42, 

'  L3x  +  22/  +  z  =  ll.  '13^-72  =  2. 

9.  4x  +  3y  =  2z  +  3.  10.    2x  +  32/  +  4z  =  17. 

11.  Find  the  number  of  the  positive  integral  solutions  of  the  equation 
3x  +  7  2/  =  1043. 

12.  Reduce  the  fraction  41/35  to  a  sum  of  two  positive  fractions 
whose  denominators  are  5  and  7. 

13.  A  man  buys  calves  at  $1  a  head  and  lambs  at  $6  a  head.  He 
spends  in  all  $110.     How  many  does  he  buy  of  each'.' 

14.  Separate  23  into  three  parts  such  that  the  sum  of  three  times  the 
first  part,  twice  the  second  part,  and  five  times  the  third  part  will  be  79. 

15.  Find  the  smallest  number  which  when  divided  by  5,  7,  9  will  give 
the  remainders  4,  6,  8. 

16.  Two  rods  of  equal  length  are  divided  into  250  and  253  equal  divi- 
sions respectively.  If  one  rod  is  laid  along  tl>e  other  so  that  their  ends 
coincide,  which  divisions  will  be  nearest  together? 


RATIO    AND    PROPORTION  347 

XIX.     RATIO   AND   PROPORTION.     VARIATION 

RATIO  AND  PROPORTION 

Ratio.     In  arithmetic  and  algebra  it  is  customary  to  extend     680 
the  use  of  the  word  ratio,  §  215,  to  numbers ;  and,  if  a  and  b 
denote  any  two  numbers,  to  define  the  ratio  of  a  to  i  as  the 
quotient  a  /b.  .  (Compare  §  216.) 

The  ratio  of  a  to  i  is  denoted  by  a /b  or  by  a:b. 

In  the  ratio  a  :  Z»  we  call  a  the  atitecedent  and  b  the  co?iseqtient. 

Properties  of  ratios.     Since  ratios  such  as  a :  6  are  fractions,     681 
their  properties  are  the  properties  of  fractions.     Hence 

The  value  of  a  ratio  is  not  clianged  when  both  of  its  terms 
are  ?nultijjlied  or  divided  by  the  same  number. 
Thus,  a  lb  =  ma  :  mb  =  a/n  :  b/n. 

On  the  other  hand,  except  when  a  =  b,  the  value  of  a:b 
is  changed  when  both  terms  are  raised  to  the  same  power,  or 
when  the  same  number  is  added  to  both.     In  particular. 

If  a,,h,  and  m  are  positive,  the  ratio  a:b  is  increased  by 

adding  m  to  both  a  and  b  wheii  a  <  b ;  decreased,  when  a  >  b. 

„  a  +  m      a      m(h  —  a) 

For  , =  ~ > 

6  +  m      b      b{b  +  m)  * 

and  7n{b  —  a)/b{b  +  ?«)  is  positive  ornegative  according  as  a<b  or  a  >b. 

Proportion.     When  the  ratios  a  :  b  and  c :  d  are  equal,  the     682 
four  numbers  a,  b,  c,  d  are  said  to  be  in  projjortion,  or  to  be 
jiroportional. 

This  proportion  may  be  written  in  any  of  the  ways 
a fb  =  c / d,  or  a\b  —  c:d,  or  a\b::c:d. 

It  is  read  ''  a  is  to  i  as  c  is  to  f/." 

In  the  proportion  a  :  b  =  c  :  d,  the  terms  a  and  d  are  called 
the  extremes,  and  b  and  c  the  means.  Again,  d  is  called  the 
fourth  proportional  to  a,  b.  and  c. 


348  A   COLLEGE   ALGEBRA 

683  Theorem.  In  any  proportion  the  product  of  the  extremes  is 
equal  to  that  of  the  means  ;  that  is, 

If  a :  b  =  c  :  d,  then  ad  =  be. 

For  from  a/b  =  c/d  we  obtain  ad  =  be  by  merely  clearing  of  fractions. 

Example.     The  first,  second,  and  fourth  terms  of  a  proportion  are 
1/2,  —  3,  and  5  respectively;  find  the  third  term. 

Calling  the  third  term  x,    l/2:-3  =  x:5. 

Hence  5-l/2=:  — 3x, 

or,  solving  for  X,  x  =  — 5/6. 

684  Conversely,  if  the  product  of  a  first  pair  of  numbers  be  equal 
to  that  of  a  second  pair,  the  four  numbers  tvillbe  in  p)ropjortion 
when  arranged  in  any  order  which  makes  one  of  the  pairs 
means  and  the  other  extremes. 

For,  let  ad  =  be. 

Dividing  both  members  by  bd,  we  have  a/b  =  c/d.  Hence 

a-.b  =:  c  :d     (1)         and         c  ■.d  =  a-.b.  (2) 

Similarly  by  dividing  both  members  of  ad  =  be  by  cd,  ab,  and  ae  in 
turn,  we  obtain 

a:c  =  b:d     (3)         and         b:d  =  a:c,  (4) 

d  -.b  =  c  -.a     (5)         and        c  la^  d  -.b,  (6) 

d:c  =  b:a     (7)         and        b:a  =  d:c.  (8) 

685  Allowable  rearrangements  of  the  terms  of  a  proportion.  From 
§§  683,  684  it  follows  that  if  a,  b,  c,  d  are  in  proportion  when 
arranged  in  any  one  of  the  orders  (l)-(8),  they  will  also  be  in 
proportion  when  arranged  in  any  other  of  these  orders.  In 
pcirticular, 

1.  In  any  2)ro}mrtion  the  terms  of  both  ratios  may  be  inter- 
changed. 

Thus,  if  a-.b  =  c\d,  then  b  :  a  =  d  :  c. 

2.  I?i  auy  j)r()j)(>rtion  either  the  means  or  the  extremes  may 
be  interchanged. 

Thus,  if  a  :  6  =  c  :  d,  then  a  :  c  =  6  :  d. 


RATIO    AND    PROPORTION  349 

The  transformations  1  and  2  are  called  inversion  and  alter- 
nation respectively. 

Other  allowable  transformations  of  a  proportion.  686 

If  we  know  that  a  :  b  =  c :  d,  we  may  conclude  that 

1.    a  +  b:b  =  c  +  d:d.  2.    a  -  b  :b  =z  c  -  d:d. 

3.    a  -\-  b  :  a  —  b  =  c  -{-  d  :  c  —  d. 
4,    ma  :  inb  =  nc  :  nd.  5.    ma  :  nb  =  mc  :  nd. 

6.    a"  :  i"  =  c"  :  d". 

For  in  1  take  the  product  of  the  means  and  extremes  and  we  have 
ad-\^bd  =  be  +  bd,  that  is  ad  =  be,  which  is  true  since  a:b  =  c  :d.  Hence 
1  is  true,  §  288.     The  truth  of  2-6  may  be  proved  in  a  similar  manner. 

The  transformations  of  a:b  =  c:  d  into  the  forms  1,  2,  3 
are  called  composition,  division,  and  comjjositio7i  and  division 
respectively. 

Example.     Solve  x2  +  2x  +  3  :  x^  _  2a;  -  3  =  2x2  +  x  -  1 :2x2-x  +  1. 
By  3,  2x2:2(2x  +  3)  =  4x2:2(x-l). 

Hence  x2  =  0,  (1) 

or  by  4,  5,  1  :  2x  +  3  =  2  :  x  -  1.  (2) 

Solving  (1)  and  (2),  x  =  0,  0,  or  -  7/3. 

Theorem.      In  a  series  of  equal  ratios  any  antecedent  is  to  its     687 
consequent  as  the  sum  of  all  the  antecedents  is  to  the  sum  of 
nil  the  consequents. 

Thus,  if  a\:bi  —  a-2  :  b-2  =  a^  :  63, 

then  Ui  :  hi  =  ai  +  a2  +  as  :  61  +  62  +  ^3. 

For  let  r  denote  the  common  value  of  the  equal  ratios.  We  then 
have 

ai/bi  =  r,  ai/b<2-r,  03/63  =  r. 

Hence  Oi  =  rhx,  a-i  =  rb-2,  «j  =  rbg, 

or,  adding,        a\  +  a^  +  as  =  r(bi  +  60  +  63). 


61  +  6a  +  &3 


350  A   COLLEGE   ALGEBRA 

Example  1.     If        x:{b  -  c)yz  =  y:{c  -  a)zx  =  z  -.{a  -h)  xy, 
then  x2  4.  y2  +  22  =  0. 

For  multiplying  the  terms  of  the  first  ratio  by  x,  those  of  the  second  by 
y,  and  those  of  the  third  by  2,  and  then  applying  our  theorem,  we  have 

X2  __  7/2  _  Z2  ^x-  +  y^  +  Z2 

{b  —  c)  xyz      (c  —  a)  xyz      {a  —  b)  xyz  0 

which  evidently  requires  that  x'^  +  y^  +  z-  =  0. 

The  device  employed  in  the  proof  just  given  will  be  found 
useful  in  dealing  with  complicated  problems  in  proportion. 

Example  2.     Prove  that  if    a  :b  =  x:y, 
then  flS  +  2  6^ :  a62  =  x3  +  2  y3  .  xy^. 

Set    a/b  =  x/y  =  r,  so  that  a  =  rb  and  x  =  ry. 

Then  (a^  +  2  5^)  /  a62  =  (rSfts  +  2  63)  /  r&3  =  (r-3  +  2)/r, 

and  (x3  +  2  y 3)  /  a;?/2  =  (r^y3  +  2  ?/3)  /  ryS  ^  (^3  +  2)  /  r. 

688         Continued  proportion.     The  numbers  a,  b,  c,  d,  ■  ■  ■  are  said  to 
be  in  continued  proportion  if  a  -.b  =  b  ic  =  c:d  :=  •  •  ■. 

If  three  numbers  a,  b,  c  are  in  continued  proportion,  so  that 
a :  b  =  b :  c,  then  b  is  called  a  mean  jnoportional  to  a  and  c, 
and  c  is  called  a  third  proportional  to  a  and  ft. 

T/'a,  b,  c  are  in  continued  proportion,  then  b^  =  ac. 
For  since  a-.b  =  b  -.c,  we  have  b-  —  ac,  §  683. 

EXERCISE  LV 

1.  Find  a  fourth  proportional  to  15,  24,  and  20;  a  third  proportional 
to  15  and  24  ;  a  mean  proportional  between  5  aW  and  20  ab^ ;  a  mean 
proportional  between  Vl2  and  V75. 

2.  IfSx  —  2y  =  x  —  5y,  find  x  :  y  ;  also  x  +  y  :  x  —  y. 

3.  If  2  x2  —  5xy  —  3  2/2  =  0,  find  x  :y;  also  y  :  x. 

4.  If  ax  +  by  +  cz  =  0  and  a'x  +  6';/  +  c'z  =  0, 
then  X  :  y  :  z  =  be'  —  b'c  :  ca'  —  c'a  :  ab'  —  a'b. 

5.  If  a  :  6  =  c  :  d,  then  06  +  cd  is  a  mean  proportional  between  a^  4-  c* 
and  62  4.  d2. 

6.  If  (a2  +  62)  cd  -  (c2  +  d2)  a^^  then  either  a:6  =  c:dora:6  =  d:C. 


RATIO    AND    PROPORTION.     VARIATION  351 


7.    Ua:b  -c:d,  then  V^  -\-Vb-Wa  +  b 


a&      63      c3         (a  +  6  +  c)^ 

9.  If  the  numbers  ai,  tto,  •••,  a„  ;  bi,  b^,  ■  ■  ■  ■,  b„  ;  ^i,  Z21  •••)  ^i  are  all  posi- 
tive, the  ratio  liai  +  hn2  +  •  •  •  +  'n««  :  ^1^1  +  ^2^2  +  •  ■  •  +  l,,bn  is  intermediate 
in  value  to  the  greatest  and  least  of  the  ratios  oi :  61,  a^ :  62,  ■  •  • ,  a„  :  6„. 

10.  li  a  —  b  -.k  =  b  —  c  :l  =  c  —  a  :  m,  and  a,  b,  c  are  unequal,  then 
k  +  I  +  m  =  0. 

11.  If  X  :  mz  —  ny  =  y  -.nx  —  Iz  =  z  -.ly  —  mx,  then  Ix  +  my  +  nz  =  0 
and  x2  +  ?/2  +  22  =  0. 

12.  If  ai :  bi  =  Uo  :  bi  =  as  :  63,  then  each  of  these  ratios  is  equal  to 

1  1 

{kal  +  Z2«2  +  '3«3)"  :  ('1^1  +  l^b^  +  kb^r- 

13.  By  aid  of  §  68G  solve  each  of  the  following  equations. 

x^  +  ax  -  a      2x2  +  rt 


(2) 


x^  —  ax  +  a      2x-  —  a 

2x3-3x2  +  2x4-2      3x3-x2  +  10x-26 


2  x"  -  3  x2  -  2  X  -  2      3  x3  -  x2  -  10  X  +  26 

14.  Separate  520  into  three  parts  in  the  ratios  2:3:5. 

15.  Two  casks  A  and  B  are  filled  with  two  kinds  of  sherry  mixed  in 
A  in  the  ratio  3  :  5,  in  B  in  the  ratio  3  :  7.  What  amount  must  be  taken 
from  each  cask  to  form  a  mixture  which  shall  consist  of  6  gallons  of  one 
kind  and  12  gallons  of  the  other  kind  ? 

VARIATION 

One  independent  variable.  If  two  variables  y  and  x  are  so 
related  that  however  their  values  may  change  their  ratio 
remains  constant,  then  y  is  said  to  vary  as  x,  or  1/  and  x  are 
said  to  vary  proportionally. 

More  briefly,  y  is  said  to  vary  as  x  when  y  / x  =  c,  ot  y  =  ex, 
where  c  denotes  a  constant. 

The  notation  y  cox  means  " y  varies  as  x." 

If  given  that  y  varies  as  x,  we  may  at  once  write  y  =  ex ; 
and  if  also  given  one  pair  of  corresponding  values  of  x  and 


352  A    COLLEGE    ALGEBRA 

y,  we  may  find  c.  The  equation  connecting  y  and  x  is  then 
known,  and  from  it  we  may  compute  the  value  of  y  which 
corresponds  to  any  given  value  of  x. 

Example.  If  y  varies  as  x,  and  y  =  12  when  x  =  2,  what  is  the  value 
of  y  when  x  =  20  ? 

We  have  y  =  ex, 

and,  by  hypothesis,  this  equation  is  satisfied  when  y  =  12,  x  =  2. 

Hence  12  =  c  •  2,  that  is  c  =  6. 

Therefore  y  =  0  x. 

Hence  when  x  =  20  we  have  y  =  (3  •  20  =  120. 

691  Instead  of  varying  as  x  itself,  y  may  vary  as  some  function 
of  x,  for  example  as  x'^,  or  as  a;  +  1,  or  as  1 1 x.  In  particular, 
if  y  varies  as  1/x,  we  say  that  y  varies  inversely  as  x. 

Example.  Given  that  y  is  the  sum  of  a  constant  and  a  term  which 
varies  inversely  as  x  ;  also  that  y  =  \  when  ic  =  —  1,  and  y  =  b  when  x  =  1. 
Find  the  equation  connecting  x  and  y. 

By  hypothesis,  y  =  a  +  b/x,  where  a  and  b  are  constants. 

Since  this  equation  is  satisfied  byx  =  -l,  y  =  l,  and  by  x  =  1,  y  =  5, 

^^  ^^^^  1  =  a  -  6  and  5  =  a  +  b. 

Hence  a  =  3,  6  =  2,  and  the  required  equation  is  y  =  3  +  2/x. 

692  More  than  one  independent  variable.  Let  a?  and  y  denote 
variables  which  are  independent  of  one  another.  If  a  third 
variable  z  varies  as  the  product  xy,  so  that  z  =  cxy,  we  say 
that  z  varies  as  x  and  y  jointly ;  and  if  z  varies  as  the  quotient 
X  /y,  so  that  «  =  c  •  x/y,  we  say  that  z  varies  directly  as  x  and 
ijiversely  as  y. 

Thus,  the  area  of  a  rectangle  varies  as  the  lengths  of  its  base  and  alti- 
tude jointly ;  and  the  length  of  the  altitude  varies  directly  as  the  area  and 
inversely  as  ihe  length  of  the  base. 

693  Theorem.  If  irlu-n  x  is  constant  z  varies  as  y,  and  token  y 
is  constant  z  varies  as  x,  then  ichen  both  x  and  y  vary,  z  varies 
as  the  product  xy. 

For  select  any  three  pairs  of  values  of  x  and  ?/,  such  as  Xi,  ?/i ;  Xn,  yi ; 
Xi,  yi ;  and  let  Zi,  Z2,  Za  denote  the  corresponding  values  of  z,  so  that 


VARIATION  363 

xi,  2/1,  zi,  (1) 

Xl,  2/2,  23,  (2) 

X2,   2/2,   Z2  (3) 

are  sets  of  corresponding  values  of  the  three  variables. 

Then  since  the  value  of  x  is  the  same  in  (1)  as  in  (2),  and,  by  hypoth- 
esis, for  any  given  value  of  x,  z  varies  as  y,  we  have,  §  689, 

21/2/1  =  23/2/2-  (4) 

Similarly  since  yn  is  common  to  (2)  and  (3),  we  have 

Zs/a-l  =22/X2.  (6) 

Multiplying  together  the  corresponding  members  of  (4)  and  (5), 

2i/a;i2/i  =  Z2/a;22/2.  (6) 

Therefore  corresponding  values  of  z  and  xy  are  proportional ;  that  is, 
z  varies  as  xy,  §  689. 

EXERCISE  LVI 

1.  If  2/  varies  as  x,  and  y  =  ~2  when  x  =  5,  what  is  the  value  of  y 
when  X  =  7  ? 

2.  If  y  varies  inversely  as  x^,  and  y  —  1  when  x  =  2,  for  what  values 
of  X  will  2/  =  3  ? 

3.  Given  that  y  is  the  sum  of  a  constant  and  a  term  which  varies  as 
X-  ;  also  that  2/  =  1  when  x  =  1,  and  y  =  0  %vhen  x  =  2.  Find  the  equation 
connecting  x  and  y. 

4.  If  y  varies  directly  as  x-  and  inversely  as  2^,  and  y  =  1  when  x  =  —  1 
and  2  =  2,  what  is  the  value  of  y  when  x  =  3  and  2  =  —  1  ? 

5.  If  y  varies  as  x,  show  that  x^  —  y-  varies  as  xy. 

6.  If  the  square  of  y  varies  as  the  cube  of  z,  and  z  varies  inversely  as 
X,  show  that  xy  varies  inversely  as  the  square  root  of  x. 

7.  The  wages  of  3  men  for  4  weeks  being  $108,  how  many  weeks  will 
5  men  work  for  $135? 

8.  The  volume  of  a  circular  disc  varies  as  its  thickness  and  the  square 
of  the  radius  of  its  face  jointly.  Two  metallic  discs  having  the  thicknesses 
8  and  2  and  the  radii  24  and  36  respectively  are  melted  and  recast  in  a 
single  disc  having  the  radius  48.     What  is  its  thickness? 

9.  A  right-circular  cone  whose  altitude  is  a  is  cut  by  a  plane  drawn 
parallel  to  its  base.  How  far  is  the  plane  from  the  vertex  of  the  cone 
when  the  area  of  the  section  is  half  that  of  the  base  ?  How  far  is  the  plane 
from  the  vertex  when  it  divides  the  cone  into  two  equivalent  parts  ? 


354  A   COLLEGE   ALGEBRA 


XX.     ARITHMETICAL    PROGRESSION 

694  Arithmetical  progression.  This  name  is  given  to  a  sequence 
of  numbers  which  may  be  derived  from  a  given  number  a  by 
repeatedly  adding  a  given  number  d,  that  is,  to  any  sequence 
which  may  be  written  in  the  form 

a,  a  +  d,  a  +  2d,  •  •  • ,  a.  +  (?i  —  1)  d.  (I) 

Since  d  is  the  difference  between  every  two  consecutive 
terms  of  (I),  it  is  called  the  common  difference  of  this  arith- 
metical progression. 

Thus,  2,  5,  8,  11  is  an  arithmetical  progression  in  which  d  =  S,  and 
2,  —  1,  —  4,  —  7  is  an  arithmetical  progression  in  which  d  =  —  ,3. 

695  The  nth  term.  Observe  that  in  (I)  the  coefficient  of  d  in  each 
of  the  terms  a,  a  +  d,  a  -]-  2  d,  ■  ■  ■  is  one  less  than  the  number 
of  the  term.  Hence  the  general  or  mih  term  is  a  -\-(m  —  l)d; 
and  if  the  entire  number  of  terms  is  n  and  we  call  the  last 
term  I,  we  have  the  formula 

l  =  a  +  (n-l)d.  (II) 

Example.  The  seventh  term  of  an  arithmetical  progression  is  15  and 
its  tenth  term  is  21 ;  find  the  first  term  a  and  the  common  difference  d ; 
and  if  the  entire  number  of  terms  is  20,  find  I. 

We  have  ^  a  +  6  d  =  15  and  a  +  9  d  =  21. 

Solving  for  a  and  d,         o  =  3,  d  =  2. 
Hence  l-S  +  19.2  =  41. 

696  The  sum.  Evidently  the  next  to  the  last  term  of  (I)  may 
be  written  I  —  d,  the  term  before  that,  I  —  2d,  ■  •  • ,  the  first 
term,  I  —(71  —  1)  d. 

Hence,  if  S  denote  the  sum  of  the  terms  of  (I),  we  have 

S  =  a+(a  +  d)  +  (a  +  2d)-] \-[a+(?i-  !)(/], 

S  =  I  -^-(l  -  d)  +  (I  -  2  d)+  ■  ■  ■  +11  -  {n  -  l)d2. 


ARITHMETICAL    PROGRESSION  355 

Adding  the  corresponding  members  of  these  two  equations, 
we  obtain  2  S  =  n(a  +  I).     Therefore 

S  =  ^(a  +  l).  (Ill) 

Example.     Find  the  sum  of  an  arithmetical  progression  of  six  terms 
■whose  first  term  is  5  and  whose  common  difference  is  4. 
Since  n  =  6,  we  have  Z  =  5  +  5  •  4  =  25. 

Hence  S  =  « (5  +  25)  =  90. 

Applications.     If  in  an  arithmetical  progression  any  three     697 
of  the  five  numbers  a,  I,  d,  n,  S  are  given,  the  formulas  (II) 
and  (III)  enable  us  to  hnd  the  other  two.     The  only  restric- 
tion on  the  given  numbers  is  that  they  be  such  as  will  lead  to 
positive  integral  values  of  >i. 


Example.     Given  d 

=  1/2, 

1  = 

3/2, 

S  =  -15/2; 

find 

a  and 

n. 

Substituting  in 

(11), 

(HI), 

3 

2  ■ 

=  a  + 

n-1 

2 

(1) 

- 

15 

'  2  ■ 

n/ 
-2V 

-D- 

(2) 

Eliminating  a, 

n2- 

-  J  n  — 

30: 

=  0. 

(3) 

Solving  (3), 

n  : 

=  10  ( 

3r  -  3. 

- 

The  value  n  = 

-.3 

is  inadmissible. 

Substituting 

;  n  = 

:10  in 

(1). 

we 

obtain  a  =  —  3.     Hence  ?i  =  10,  a  =  —  3,  and  the  arithmetical  progression 

is  -  .3,  -  2.V,  -  2,  -  li,  -  1,  -  i,  0,  I,  1,  11. 

Arithmetical  means.     If  three  numbers  form  an  arithmetical     698 
progression,  the  middle  number  is  called  the  arithmetical  mean 
of  the  other  two. 

The  arithmetical  mean  of  any  two  numbers  a  and  b  is  one 
half  of  their  sum. 

For  if  X  be  the  arithmetical  mean  of  a  and  6,  then  the  sequence  a,  x,  b 
is  an  arithmetical  progression. 

Hence  x  —  a  —  b  —  x, 

and  therefore  x  —  {a  +  b)/2. 


356  A   COLLEGE    ALGEBRA 

In  any  arithmetical  progression  all  the  intermediate  terms 
may  be  called  the  arithmetical  means  of  the  first  and  last 
terms.  It  is  always  possible  to  insert  or  "  interpolate  "  any 
number  of  such  means  between  two  given  numbers  a  and  h. 

Example.     Interpolate  four  arithmetical  means  between  3  and  5. 

We  are  asked  to  find  the  intermediate  terms  of  an  arithmetical  progres- 
sion in  which  a  =  3,  I  —  b,  and  ?i  =  4  +  2  or  6. 

Substituting  i  =  5,  a  =  3,  ?i  =  6  in  (II),  we  have 
5  =  3  +  5  d,  whence  d  =  2/5. 

Hence  the  required  means  are  3§,  3f ,  4i,  4f. 

EXERCISE   LVII 

1.  Find  the  twentieth  term  and  the  sum  of  the  first  twenty  terms  of 
3,  6,  9,  •••;  of  -3,  -1\,  0,  •  •  • . 

2.  Find  a  formula  for  the  sum  to  n  terms  of  1,  2,  3,  •  •  • ;  of  1,  3,  5,  •  •  • ; 
of  2,  4,  G,  ••-. 

3.  Find  the  sum  of  the  first  n  numbers  of  the  form  6r  +  1,  where  r 
denotes  0  or  a  positive  integer. 

4.  Find  the  arithmetical  progression  of  ten  terms  whose  fifth  term  is 
1  and  whose  eighth  term  is  2. 

5.  Insert  five  arithmetical  means  between  —  1  and  2 

6.  Given  n  =  IG,  a  =  0,  d  =  4/3  ;  find  Z  and  S. 

7.  Given  n  =  7,  Z  =  -  7,  (Z  =  -  5/3  ;  find  a  and  S. 

8.  Given  n  =  12,  a  =  -  5/3,  Z  =  31 1 ;  find  d  and  S. 

9.  Given  a  =  2,  Z  =  -  23^,  .S  =  -  559  ;  find  n  and  d. 

10.  Given  n  =  7,  n  =  3/7,  .S  =  45  ;  find  d  and  /. 

11.  Given  a  =  4,  cZ  =  1/5,  Z  =  9s  ;  find  n  and  S. 

12.  Given  n  =  9,  d  =  -  4,  .S  =  135  ;  find  a  and  I. 

13.  Given  n  =  10,  I  =  -  2,  S  =  Ui>;  find  a  and  d. 

14.  Given  d  =  5,  Z  =  -  47,  S  =  -  357  ;  find  n  and  a. 

15.  Given  a  =  -  10,  d  =  7,  .S  =  20  ;  find  n  and  I. 

16.  Show  Ihat  if  «'-,  h'-,  c-  are  in  arithmetical  progression,  so  also  are 
1/(6 +  c),  l/(c  +  «),  1/(a  +  b). 


GEOMETRICAL    PROGRESSION  357 

17.  Show  that  the  sum  of  any  n  consecutive  integers  is  divisible  by  n, 
if  n  be  odd. 

18.  Find  an  arithmetical  progression  such  that  the  sum  of  the  first  three 
terms  is  one  half  the  sum  of  the  next  four  terms,  the  first  term  being  1. 

19.  Three  numbers  are  in  arithmetical  progression.  Their  sum  is  15 
and  the  sum  of  their  squares  is  83.     Find  these  numbers. 

20.  Find  the  sum  of  all  positive  integers  of  three  digits  which  are 
multiples  of  9. 

21.  If  a  person  saves  $\oO  a  year  and  at  the  end  of  the  year  puts  this 
sum  at  simple  interest  at  4%,  to  how  much  will  his  savings  amount  at  the 
end  of  11  years? 

22.  Two  men  A  and  B  set  out  at  the  same  time  from  two  places  72 
miles  apart  to  walk  toward  one  another.  If  A  walks  at  the  rate  of  4 
miles  an  hour,  while  B  walks  2  miles  the  first  hour,  2?^  miles  the  second 
hour,  3  miles  the  third  hour,  and  so  on,  when  and  where  will  they  meet  ? 


XXI.     GEOMETRICAL    PROGRESSION 

Geometrical  progression.     This  name  is  given  to  a  sequence     699 
of  numbers  which  may  be  derived  from  a  given  number  a  by 
repeatedly  multiplying  by  a  given  number  r,  that  is,  to  any 
sequence  which  may  be  written  in  the  form 

a,  ar,  ar'^,  •■-,  ar"~^.  (I) 

We  call  r  the  common  ratio  of  the  geometrical  progression 
(I)  and  say  that  the  progression  is  increasing  or  decreasing 
according  as  r  is  numerically  greater  or  less  than  1. 

Thus,  1,  2,  4,  8  and  1,  -  2,  4,  -  8  are  increasing  geometrical  progres- 
sions in  which  r  =  2  and  —  2  respectively;  while  1,  1/2,  1/4,  1/8  is  a 
decreasing  geometrical  progression  in  which  r  —  \  /2. 

The  nth  term.     Observe  that  the  exponent  of  r  in  each  term     700 
of  (I)  is  one  less  than  the  number  of  the  term.     Hence  in  a 
geometrical  progression  of  n  terms  whose  first  term  is  a  and 
whose  ratio  is  r,  the  formula  for  the  last  term  I  is 

I  =  ar-\  (II) 


358  A    COLLEGE    ALGEBRA 

701  The  sum.  Let  S  denote  the  sum  of  the  geometrical  pro- 
gression (I). 

Then  S  =  a -{- ar  +  ar^ -] +  ar"-^  +  ar"-^ 

and  rS  =  ar  -{-  ar^  +  •  •  •  +  a?*""^  +  ar"~^  +  ar\ 

Subtracting  the  second  of  these  equations  from  the  first,  we 
obtain  (1  —  r)S  =  a  —  ar'\     Therefore 

^^a^,-^  (III) 

In  applying  this  formula  to  an  increasing  geometrical  pro- 
gression we  may  more  conveniently  write  it  thus: 

S  =  a{i^-l)/ir-l). 

From  (II)  we  obtain  rl  =  ar".     Hence  (III)  may  also  be 
written  thus  :  S  =  (a-rl)/{l  -  r),  or  5  =  (rl  -  a)  /  (r  -  1). 

Example.     In  the  geometrical  progression  2,  —  4,  8,  —  16,  •  •  •  to  eight 
terms,  find  I  and  S. 

Here  a  =  2,  r  =  —  2,  and  n  =  8." 

Hence,  by  (II),  l  =  2{-2y  =  -  256, 

..,.„„,,  «  =  .L^._-. 

702  Applications.  If  in  a  geometrical  progression  any  three  of 
the  five  numbers  a,  I,  r,  n,  S  are  given,  the  formulas  (II)  and 
(III)  determine  the  other  two.  Moreover  these  two  numbers 
can  be  actually  found  by  methods  already  explained,  except 
when  the  given  numbers  are  a,  n,  S  or  I,  n,  S.  If  one  of  the 
unknown  numbers  is  n,  it  must  be  found  by  inspection ;  but 
this  is  always  possible  if  admissible  values  have  been  assigned 
to  the  given  numbers,  since  n  will  then  be  a  positive  integer. 

Example  1.     Given  r  =  3,  n  =  6,  S  =  728  ;  find  a  and  I. 
Substituting  the  given  values  in  (II)  and  (III),  we  have 

i  =  a  •  35  =:  243  a,  and  728  =  a^-^  =  364a. 
3-1 

Solving  these  equations,       a  =  2,  1  =  486. 


GEOMETRICAL    PROGRESSIOX  359 

Example  2.     Given    a  =  6,  n  =  5,  1  =  2/27  ;  find  r  and  S. 

By  (II),  2/27  =  6r*,  whence  r*  =  1/81,  or  r  =  ±  1/3. 

Therefore,  by  (III),  if  r  =  1  /3,  then  S  =  6  ^-(^/^)^  =  ^ , 
'    "^^      ''  1-1/3         27 

and  if  r  =  -l/3,  then  S  =  e^-llil/^  =  '^. 

l-(-l/3)        27 

Hence  there  are  two  geometrical  progressions  in  which  a  =  6,  n  =  5, 
and  i  =  2/27. 

Example  3.     Given  a  =  -  3,  Z  =  -  46875,  S  =  -  39063  ;  find  r  and  n. 
Substituting  in  the  formula  S  =  (a  —  rl) /  (I  —  r),  §  701,  we  have 

_  39063  =  -  3 +  46875  r^  ^^^^^^  r  =  -b. 
1  —  r 
Therefore,  by  (II),  _  46875  =  - 3(- 5)"-i,  or  (- 5)"-!  =  15625. 
But  by  factoring  15625  we  find  15625  =  5^  =  (-  5)6. 
Hence  n  —  1  =  6,  that  is  n  =  7. 

Example  4.     Given  a  =  3,  n  =  5,  -S  =  93  ;  find  r  and  I. 

By  (III),  93  =  3^^^  =  3 (H-  r  +  r2  +  r3  +  ^4). 

Hence  r*  +  r*  +  r^  +  r  -  30  =  0. 

Thus  this  problem  involves  solving  an  equation  of  the  fourth  degree ; 
and,  in  general,  when  a,  n,  S  are  the  given  numbers,  to  find  r  we  must 
solve  an  equation  of  the  degree  n  —  1.  In  this  particular  case,  however, 
we  may  find  one  value  of  r  by  the  method  of  §455.     It  is  2. 

Substituting  r  =  2  in  (II),  we  have  Z  =  3  •  2*  =  48. 

Geometrical  means.     If  three  numbers   form  a  geometrical     703 
progression,  the  middle  number  is  called  the  geometrical  mean 
of  the  other  two. 

The  geometrical  mean  of  any  two  numbers  a  and  b  is  a 
square  root  of  their  x>rod;uct. 

For  if  X  be  the  geometrical  mean  of  a  and  6,  the  sequence  a,  x,  h  is 
a  geometrical  progression. 

Hence  x/a  =  h/x  and  therefore  x  =  ±  Va6. 

In  any  geometrical  progression  all  the  intermediate  terms 
may  be  called  the  geometrical  means  of  the  first  and  last  terms. 


360  A    COLLEGE    ALGEBRA 

We  may  insert  any  number  of  such  means  between  two  given 
numbers  a  and  b,  as  in  the  following  example. 

Example.     Insert  four  geometrical  means  between  18  and  2/27. 
It  is  required  to  find  the  intermediate  terms  of  a  geometrical  progres- 
sion in  which  a  =  18,  i  =  2/27,  and  n  =  4  -f  2  =  6. 
Substituting  the  given  values  in  (II),  we  have 

2/27  =  18 r5,  whence  r  =  1/3. 
Hence  the  means  are  6,  2,  2/3,  2/9. 

704         Infinite  decreasing  geometric  series.     We  call  an  expression  of 
the  form  ^  ^  ^^.  ^  ^,.,  ^  . . .  _^  ^,.„_i  _^  . .  .^  ^^^ 

supposed  continued  without  end,  an  infinite  geometric  series. 

By  the  formula  (III),  the  sum  of  the  first  n  terms  of  (1)  is 
a{l-r")/(l~r). 

Suppose  that  ?•  <  1  numerically.  Then,  as  n  is  indefinitely 
increased,  y"  will  approach  0  as  limit,  §  724,  and  therefore 
a(l  —  ?•")/(!  —  /•)  will  approach  a /(I  —  r)  as  limit.  We  call 
this  limit  the  sum  of  the  infinite  series  (1).  Hence,  if  S  denote 
the  sum  of  (1),  we  have 

Example  1.     Find  the  sum  of  1  +  1/2  -l-  1/4  +  1/8  H . 

Here  a  =  1  and  r  =  1  /2. 

Hence  S  =  1/[1  -  1/2]  =  2. 

Example  2.     Find  the  value  of  the  recurring  decimal  .72323  ■  •  • . 

2.3  23 

The  part  wliich  recurs  may  be  written 1 1 ,  and,  by  (2), 

1000      100000  '    J  V  /' 

023  23 

the  sum  of  this  infinite  series  is  -^—^ —  or Adding  .7,  the  part  which 

l-.Ol       990  &     .        1 

358 

does  not  recur,  we  obtain  for  the  value  of  the  given  decimal 

495 

EXERCISE   LVIII 

1.  Find  the  fifth  term  and  the  sum  of  the  first  five  terms  of  the 
geometrical  progre.ssion  2,    —  6,   18,  ■  •  ■. 

2.  Find  the  fourth  term  and  the  sum  of  the  first  four  terms  of  the 
geometrical  progression  4,  6,  9,  •  •  •. 


GEOMETRICAL    PROGRESSION  361 

3.  Find  the  sums  of  the  following  infinite  geometric  series : 

12_6+3-...;l-i  +  i----;f  +  i+i^  +  ---. 

4.  Find  the  values  of  the  following  recurring  decimals : 

.341341  ••-,  .0567272...,  8.45164516  ••  =. 

5.  Given  a  =  -  .03,  r  =  10,  n  =  6  ;  find  I  and  *\ 

6.  Given  n  =  7,  a  =  48,  i  =  3/4  ;  find  r  and  S. 

7.  Given  a  =  1/16,  r  =  2,  Z  =  8  ;  find  n  and  S. 

8.  Given  n  =  5,  r  =  -3,  l  =  Sl;  find  a  and  S. 

9.  Given  a  =  54,  r  =  1/3,  6'  =  80| ;  find  n  and  I. 

10.  Given  n  =  4,  a  =  -  3,  S  =  -  408 ;  find  r  and  I. 

11.  Given  a  =-9/16,  Z  =  -16/9,  S  =  - 781/144;  find  n  and  r. 

12.  Given  n  =  0,  r  =  -  2/3,  S  =  665/216  ;  find  a  and  I. 

13.  Given  r  =  3/2,  I  =  mi,  S  =  83J  ;  find  n  and  a. 

14.  Given  7i  =  4,  Z  =  54/25,  S  =  544/25 ;  find  a  and  r. 

15.  Given  n  =  5,  Z  =  48,  S  =  93 ;  find  a  and  r. 

16.  Find  the  positive  geometrical  mean  of  0^/6  and  h^/a. 

17.  Insert  three  geometrical  means  between  5  and  405. 

18.  The  third  term  of  a  geometrical  progression  is  3  and  the  sixth 
term  is  —  3/8.     Find  the  seventh  term. 

19.  Find  a  geometrical  progression  of  four  terms  in  which  the  sum  of 
the  first  and  last  terms  is  133  and  the  sum  of  the  middle  terms  is  70. 

20.  Find  three  numbers  in  geometrical  progi'ession  such  that  their 
sum  shall  be  7  and  the  sum  of  their  squares  91. 

21.  Three  numbers  whose  sum  is  36  are  in  arithmetical  progression. 
If  1,  4,  43  be  added  to  them  respectively,  the  results  are  in  geometrical 
progression.     Find  the  numbers. 

22.  There  are  four  numbers  the  first  three  of  which  are  in  arithmetical 
progression  and  the  last  three  in  geometrical  progression.  The  sum  of 
the  first  and  fourth  is  16  and  the  sum  of  the  second  and  third  is  8.  Find 
the  numbers. 

23.  What  distance  will  an  elastic  ball  traverse  before  coming  to  rest  if 
it  be  dropped  from  a  height  of  15  feet  and  if  after  each  fall  it  rebounds 
to  2/3  the  height  from  which  it  falls? 


362  A   COLLEGE   ALGEBRA 

XXII.     HARMONICAL   PROGRESSION 

705  Harmonical  progression.  This  name  is  given  to  a  sequence  of 
numbers  whose  reciprocals  form  an  arithmetical  progression, 
that  is,  to  any  sequence  which  may  be  written  in  the  form 

1/a,   \l{a^d),   l/(a  +  2fi),  •••,  l/[a+(;i-l)rf]. 

Thus,  1,  1/2,  1/3,  1/4  and  3/2,  3/4,  3/6,  3/8  are  harmonical 
progressions. 

Example.  Prove  that  if  a,  6,  c  are  in  harmonical  progression,  then 
a:c=ia  —  6:6  —  c. 

Since  1/a,  1/6,  1/c  is  an  arithmetical  progression,  we  have 

l/6-l/a  =  1/c-  1/6. 
Hence  c  (a  —  6)  =  a  (6  —  c),  that  is  a  :  c  =  a  —  6  :  6  —  c. 

706  To  find  any  particular  term  of  an  harmonical  progression, 
we  obtain  the  term  which  occupies  the  same  position  in  the 
corresponding  arithmetical  progression  and  invert  it. 

Example.  Find  the  tenth  term  of  the  harmonical  progression  3/5, 
3/7,  3/9,  •■■. 

By  §  695,  the  tenth  term  of  the  corresponding  arithmetical  progression 
5/3,  7/3,  9/3,  ••  •  is  23/3.  Hence  the  tenth  term  of  3/5,  3/7,  3/9,  ••• 
is  3/23. 

707  Harmonical  means.  If  three  numbers  are  in  harmonical  pro- 
gression, the  middle  number  is  called  the  harmonical  mean 
of  the  other  two.  Again,  in  any  harmonical  progression  all 
the  intermediate  terms  may  be  called  the  harmonical  means 
of  the  extreme  terms. 

Example  1.     Find  the  harmonical  mean  of  a  and  6. 
If  this  mean  be  x,  then  1  /a,  1  /x,  1  /6  is  an  arithmetical  progression. 
Hence       l/x  -  1/a  =  1/6  -  1/x,  or  2/x  =  1/a  +  1/6. 
Therefore  x  =  2  a6/  (a  +  6). 

Example  2.  Prove  that  the  geometrical  mean  of  two  numbers  a  and  6 
is  also  the  geometrical  mean  of  their  arithmetical  and  harmonical  means. 

Let  A,  G,  and  ZT denote  respectively  the  arithmetical,  geometrical,  and 
harmonical  means  of  a  and  b. 


HARMONICAL    PROGRESSION  363 


Then  ^  =  ^+^,  G  =  V^,  iT^li^. 

2  a  +  b 

J  rr       «  +  ft     2  a6  ,        ^, 

Henco  AH  = =  ab  =  G^. 

2       a  +  6 

Example  3.     Prove  that  when  a  and  6  are  positive,  A>  G>H. 

„,    ,                  .       „      a  +  b       2ab        {a-b^ 
We  have  A  -  H  = =  -i '—. 

2  a  +  b      2  {a  +  b) 

Therefore,  since  (a  -  6)V2 (a  +  6)  is  positive,  we  have  A>H. 
And  since,  by  Ex.  2,  G  is  intermediate  in  value  to  A  and  H,  we  have 
A>G>H. 

EXERCISE  LIX 

1.  Continue  the  harmonical  progression  3/5,  3/7,  1/3  for  two 
terms. 

2.  Find  the  harmonical  mean  of  3/4  and  5. 

3.  Insert  four  harmonical  means  between  10  and  15. 

4.  The  second  and  fourth  terms  of  an  harmonical  progression  are  4/5 
and  —  4.     Find  the  third  term. 

5.  The  arithmetical  mean  of  two  numbers  is  4  and  their  harmonical 
mean  is  15/4.     Find  the  numbers. 

6.  The  geometrical  mean  of  two  numbers  is  4  and  their  harmonical 
mean  is  16/5.     Find  the  numbers. 

7.  Show  that  if  a,  b,  c  are  in  harmonical  progression,  so  also  are 
tt/(6  +  c),  b/{c  +  a),  c/{a  +  b). 

8.  Three  numbers  are  in  harmonical  progression.  Show  that  if  half 
of  the  middle  term  be  subtracted  from  each,  the  results  will  be  in  geomet- 
rical progression. 

9.  Show  that  if  x  is  the  harmonical  mean  between  a  and  6,  then 
l/(x-a)  +  \/{x-b)  =  \/a  +  l/b. 

10.  The  bisector  of  the  vertical  angle  C  of  the  triangle  ABC  meets 
the  base  AB  at  D,  and  the  bisector  of  the  exterior  angle  at  C  meets  AB 
produced  at  E.     Show  that  AD,  AB,  AE  are  in  harmonical  progression. 

11.  The  point  P  lies  outside  of  a  circle  who.se  center  is  0,  and  the 
tangents  from  P  touch  the  circle  at  T  and  T.  If  tlie  line  PO  meets  the 
circle  at  A  and  B  and  TT  at  C,  show  that  PC  is  the  harmonical  mean 
between  PA  and  PB. 


364  A  COLLEGE   ALGEBRA 

XXIII.     METHOD    OF   DIFFERENCES.     ARITH- 
METICAL  PROGRESSIONS    OF    HIGHER 
ORDERS.     INTERPOLATION 

ARITHMETICAL  PROGRESSIONS   OF  HIGHER  ORDERS 

708  Differences  of  various  orders.  If  in  any  given  sequence  of 
numbers  we  subtract  each  term  from  the  next  following  term, 
we  obtain  a  sequence  called  the  first  order  of  differences  of  the 
given  sequence;  if  we  treat  this  new  sequence  in  a  similar 
manner,  we  obtain  the  second  order  of  differences  of  the  given 
sequence  ;  and  so  on. 

Thus,  if  the  given  sequence  be  1^,  2\  .S^,  •  •  • ,  we  have 

given  sequence  1,     8,  27,  64,  125,  216,  •  •  • , 

first  differences  7,  19,  37,  61,     91,  ••-, 

second  differences        12,  18,  24,  30,  •  •  • , 
third  differences  6,     6,     6,  •  •  • . 

The  fourth  and  all  subsequent  differences  are  0. 

709  Arithmetical  progression  of  the  rth  order.  This  name  is  given 
to  a  sequence  whose  rth  differences  are  equal,  and  whose 
subsequent  differences  are  therefore  0. 

Thus,  1^,  2^,  33,  4-^,  •  •  •  is  an  arithmetical  progression  of  the  third  order, 
for,  as  just  shown,  its  third  differences  are  equal. 

An  ordinary  arithmetical  progression,  §  694,  is  of  the  first  order,  each 
of  its  first  differences  being  the  common  difference  d. 

710  The  nth  term  of  an  arithmetical  progression  of  the  rth  order. 

Given  any  arithmetical  progression  of  the  rth  order 

«1,     ^2,     «3J     ^4,    •••»     «„,     «„  +  !'    '■■,  (1) 

and  let  d^,  d^,  ■■■,  d,.  denote  the  first  terms  of  its  successive 
orders  of  differences.  We  are  to  obtain  a  formula  for  a„  in 
terms  of  a^,  r/,,  d«,  ■■■,  d,.,  and  71. 

The  first  order  of  differences  of  (1)  is 


METHOD    OF   DIFFERENCES  365 

The  first  term  of  (2)  is  di,  and  the  first  terms  of  its  first, 
second,  •  •  •  differences  are  iL,  d^,  ■  •  • ;  for  the  first  differences 
of  (2)  are  the  second  differences  of  (1),  and  so  on. 

Hence  when  we  have  found  an  expression  for  any  particular 
term  of  (1),  we  can  derive  from  it  an  expression  for  the  corre- 
sponding term  of  (2)  by  applying  the  rule  : 

Replace  a^,  d^,  d^,  •  •  •  by  d^,  d^,  d^  ••■.  (3) 

Now  since  d^  —  a»  —  a^,  we  have  a^  =  Oi  -{-  dy.  Starting 
with  this  formula  for  a^,  we  may  reckon  out  a 3,  04,  •  •  •  as 
follows : 


We  have 

aj  =  ai  +     dy 

Hence,  by  (3), 

^3  - 

-  a2  =                     C?i  +       (^2 

Adding, 

^3  =  f/i  +  2  r/i  +      f/a 

Hence,  by  (3), 

04- 

-«3=                        dy   +  2d._-\-d^ 

Adding, 

a^  =  ay  +  3(/i  +  3f/o  +  (/3 

and  so  on  indefinitely,  the  reckoning,  so  far  as  coefficients  are 
concerned,  being  precisely  the  same  as  that  given  in  §  311  for 
finding  the  coefficients  of  successive  powers  of  a  +  h.  There- 
fore, by  §  561,  we  have  the  formula 

««  =  «i  +  ("  —  l)<^^i 

^  (..-l)(..-2)  ,^  ^  . . .  +  (.-^V..(.-.)^    ^j^ 

Example.     Compute  the  fifteenth  term  of  1^,  2-^,  •  •  •  by  this  formula. 
Here,  §  708,       a^  =  1,  (h  =  7,  d.  =  12,  ds  =  dr  =  G. 

14    13              14  ■  1.3  •  12 
Hence  ais  =  1  +  14  ■  7  +  — ^j—  ■  12  +  — -^-^ 6  =  3375. 

Sum  of  the  first  n  terms  of  an  arithmetical  progression  of  the     711 
rth  order.     Let  S„  denote  this  sum,  the  sequence  being 

«!,   aa,   as,  ■■-,  a„,   a„^j,  •••,  (1) 

and  di,  d^,---,  d^  having  the  same  meanings  as  in  §  710. 


366  A   COLLEGE   ALGEBRA 

Form  the  sequence  of  which  (1)  is  the  first  order  of  differ- 
ences, namely : 

0,     «!,     rti   +   «2)     «1   +  «2   +  ^3,  •  •  •,     «!   +  «2   +   •  •  •   +  Cini  '  '  "•     (2) 

Then  S„  is  the  (?i  +  l)th  term  of  (2),  and  since  (2)  is  an 
arithmetical  progression  of  the  («  +  l)th  order  whose  first 
term  is  0,  and  the  first  terms  of  its  several  orders  of  differences 
are  «i,  di,  cL,  ■  ■  ■  d,.,  we  have,  by  (I), 

Example.     Find  the  sum  of  the  first  fifteen  terms  of  1^,  2^,  3^,  •  •  •. 

Here,  §  708,  n  =  15,  ai  =  1,  ch  =  7,  dz  =  12,  ds  =  d,.  =  6. 

nr     15-14    „      15    14    13    ,„      15    14    13    12    ^     ,,,^^ 

Hence  Si5  =  15+ 7  + 12  + 0  =  14400. 

2  2-3  2.3-4 

712         Piles  of  spherical  shot.     1.    To  find  the  number  of  shot  when 
the  pile  has  the  form  of  a  triangular  pyramid. 

The  top  course  contains  1  shot,  the  next  lower  course  1  +  2  shot,  the 
next  1  +  2  +  3,  and  so  on. 

Hence,  if  there  are  n  courses,  the  number  of  shot  is  the  sum  of  n  terms 
of  the  sequence  1,  3,  6,  10,  15,  •  •  •. 

The  first  differences  of  this  sequence  are  2,  3,  4,  5,  •  •  • ,  and  the  second 
differences  are  1,  1,  1,  •  •  • . 

Hence  1,  3,  6,  ■  •  •  is  an  arithmetical  progression  of  the  second  order  in 
which  ai  =  1,  di  =  2,  (fj  =  1. 

Therefore,  by  (11),  .S„  =  n  +  ''S'l^l^  .  2  +  n(n~l){n-2) 
1-2  1.2.3 

_  n  (n  +  1)  (n  +  2) 
1.2-3 
Thus,  in  a  pile  of  twenty  courses  there  are  20  .  21  .  22/6  =  1540  shot. 
2.    To  find  the  number  of  shot  when  the  pile  has  the  form 
of  a  pyramid  with  a  square  base. 

Enumerating  the  shot  by  courses  as  before,  we  obtain  the  sequence 
P,  22,  32,  42,  ... . 

The  first  differences  are  3,  5,  7,  . .  • ,  and  the  second  are  2,  2,  ■  •  • . 

Hence  12,  22,  32,  ...  is  an  arithmetical  progression  of  the  second  order 
in  which  ui  =  l,di  =  3,  dz  =  2. 


METHOD   OF   DIFFERENCES  367 

Therefore, by (II),  5„:=n  +  ^^-3  +  ^<--;^^(;-^).2 

^n(n  +  l)C2n  +  l). 

1-2-3 

Thus,  when  n  =  20,  the  pile  contains  20  •  21  •  41/6  =  2870  shot. 

3.  To  find  the  number  of  shot  when  the  pile  has  a  rectan- 
gular base  and  terminates  at  the  top  in  a  row  of  ^^  shot. 

Again  enumerating  the  shot  by  courses,  we  obtain  the  sequence 
p,  2{p+l),  3(p  +  2),  4(p  +  3),.... 

The  first  differences  are  p  +  2,  p  +  i,  p  +  6,  ■  ■  ■ ,  and  the  second  differ- 
ences are  2,  2,  •  •  • . 

Hence  p,  2  (p  +  1),  3  (p  +  2),  •  •  •  is  an  arithmetical  progression  of  the 
second  order  in  which  ai  =  p,  di  —  p  +  2,  and  dz  =  2. 

Therefore,  by  (II),  -S„  =  np  +  "^^^  (p  +  2)  +  ^n -IHn^- 2)  _  ^ 
_n{n  +  l){Sp  +  2n-2) 

Thus,  when  ?i  =  20  andp  =  5,  the  number  of  shot  is  20  •  21  •  53/6  =  3710. 

A  theorem  respecting  arithmetical  progressions.  An  examina-  713 
tion  of  the  formula  for  the  nth  term  of  an  arithmetical 
progression  of  the  rth  order,  §  710,  (I),  will  show  that  if  we 
carry  out  the  indicated  multiplications  and  arrange  the  result 
according  to  descending  powers  of  n,  we  can  reduce  it  to  the 
form 

where  the  coefficients  b^,  b^,  ■•  -,  b^  are  independent  of  n. 

Thus,  when  r  =  2,  we  have 

(n-l)(n-2), 
a„  =  ai  +  (n  -  1)  di  +  ^^ ^2 

=  I n2+  {di  -  ^ d2) n  +  (ai  -  d,  +  d^). 

Therefore  the  terms  of  any  arithmetical  progression  of 
the  rth  order,  Oj,  a^,  a^,  ■  ■  ■,  are  the  values  for  71  =  1,  2,  3,  •  ■  ■ 
of  a  certain  polynomial  b^n''  +  bin'"^  +  ■  ■  ■  +  b^  whose  degree 
with  respect  to  n  is  r.     We  are  going  to  show  conversely  that 


368  A   COLLEGE    ALGEBRA 

714         Theorem.     If  <f>(x.)  denote  any  rational  integral  function  of 
the  vth  degree,  as 

the  sequence  of  numbers  ^(1),  <^(2),  <^(3),  •••,  obtained  by 
setting  x  =  1,  2,  3,  •  •  •  successively  in  <^(x),  is  an  arithmetical 
progression  of  the  rth  ojrler. 

Here  the  given  sequence  of  numbers  is 

</.(!),  c^ (2),  <^ (3),  <^ (4),...  (1) 

and  we  are  to  prove  that  all  of  its  rth  differences  are  equal. 
Evidently  the  first  differences  of  (1),  namely 

.^(2)- </>(!),  <^(3)-<^(2),  cA(4)-<^(3)...,  (2) 

are  the  values  of  ^(x  +  1)  —  <)!> (x)  for  a;  =  1,  2,  3,  •  •  •. 

But  <^(.r  +  1)  —  4>(^)  '^i^y  be  reduced  to  the  form  of  a  poly- 
nomial in  X.     Call  this  polynomial  <^i  (x).     Its  degree  is  r  —  1. 

For,  by  the  binomial  theorem,  §  561,  we  have 

<p{x  +  l)-<p{x)  =  6o(x  +  ])'-  +  6i(x  +  l)'-i  + (6oX'-  +  M'-i  +  ---) 

=  boX'-  +  rboX'--'^  +  - ■■  +  biX'--'^-\ (?>or'"+ 6iX''-i  + •    •) 

::=  rboX'-^  +  •■■ 

Similarly,  if  we  write 

<^l  (X  +  1)  -  </>!  (X)  =  <f>2  (X),     <^2  (X  +  1)  -  <^2  (X)  =  (f>Z  (X), 

and  so  on,  the  values  of  ^3(x),  cf>3(x),  ■•■,  <f>r(x)  for  x  =  1, 
2,  3,  ■   ■  will  be  the  second,  third,  •  •  •,  rth  differences  of  (1). 

But  <^,.(x)  is  a  constant  and  the  rth  differences  of  (1)  are 
therefore  equal.  For  the  degree  of  ^gC^)  is  (r  — 1)—  1,  or 
r  — 2;  that  of  <f>3(x)  is  r  — 3;  and  finally  that  of  ^^(x)  is 
r  —  r,  or  0. 

For  example,  if  0  (x)  =  2  x^  —  x  +  1,  we  have 
01  (x)  ==  2  (X  +  1)3  -  (X  +  1)  +  1  -  (2  x3  -  X  +  1)  =  6x2  +  6x  +  1, 
02(x)  =  6(x  +  l)2  +  6(x  +  1)  +  1  -  (Gx2  +  'jx  +  l)  =  12x  +  12, 
<pa{x)  =  12(x  +  l)  +  12-(12x  +  12)  =  12. 


METHOD    OF    DIFFERENCES  369 

Hence  the  values  of  6x'^  +  6x  +  l,  12 x  +  12,  12  for  x  =  1,  2,  3,  •  • . 
are  the  first,  second,  third  differences  of  the  corresponding  values  of 
2x'  —  X  +  1  ;  and  the  third  differences  are  equal,  all  being  12. 

Thus,  for  X  =  1,  2,  3,  4,  5,  •  •  • ,  we  find 


2x3-x  +  l=    2,   15,  62,   125,  240,- 


(1) 


(2) 
(3) 
(4) 

(3),  (4)  actually 


6x2  +  Ox +  1  =  13,  37^  73^  121,  181, 
12  X  +  12  =24,  36,  48,     60,     72, 
12  =  12,   12,   12,     12,     12, 
And  by  comparing  (1),  (2),  (3),  (4),  we  find  that  (2), 
are,  as  they  should  be,  the  first,  second,  tliird  differences  of  (1). 

Corollary  1.      The  vth  powers  of  consecutioe  integers  form  an     715 
arithmetical  progression  of  the  xth  order. 

For  l*",  2'',  3'",  •  ■  ■  are  the  values  of  the  rational  integral  function  of  the 
rth  degree  0  (x)  =  x''  for  x  =  1,  2,  3,  •  •  • . 

Corollary  2.      The  2^^'<^ducts  of  the  corresponding  terms  of  two     716 
arithmetical  j^^'ogressions,   the   one   of  the  rth  order  and  the 
other  of  the   sth  order,  form  an  arithmetical  progression  of 
the  (r  +  &)th  order. 

For  the  product  of  a  rational  integral  function  of  the  rth  degree  by  one 
of  the  sth  degree  is  a  rational  integral  function  of  the  (r  +  s)th  degree. 

EXERCISE  LX 

1.  Find  the  twentieth  term  and  the  sum  of  the  first  twenty  terms  of 
the  sequence  1,  2,  4,  7,  •  •  •. 

2.  Find  the  eightieth  term  and  the  sum  of  the  first  eighty  terms  of 
the  sequence  3,  8,  15,  24,  35,  •  •  •. 

3.  Determine  the  order  of  each  of  the  following  arithmetical  pro- 
gressions. 

(1)  3,  0,   -  1,  0,  3,  •••,  (2)  10,  38,  88,  166,  278,  430,  •••, 

(3)  285,  204,  140,  91,  55,  •••,     (4)  2,  20,  90,  272,  650,  1332,  •••. 
Also  find  the  eighteenth  term  of  (1),  the  twentieth  term  of  (2),  the 
twelfth  term  of  (3),  and  the  tenth  term  of  (4). 

4.  What  is  the  order  of  1  •  2  •  3,  2  ■  3  •  4,  3  •  4  •  5,  •  ■  ■  ?     What  is  its  nth 
term  ?   the  sum  of  its  first  n  terms  ? 

What  is  the  order  and  what  the  nth  term  of  1  •  4  •  22,  2  ■  6  •  3^,  3  •  8  •  42,  ■ . .  ? 


370  A   COLLEGE   ALGEBRA 

5.  Find  the  number  of  shot  in  a  triangular  pile  of  fourteen  courses. 
How  many  shot  are  there  in  the  lowest  course? 

6.  If  from  a  square  pile  of  fifteen  courses  six  courses  be  removed, 
how  many  shot  remain  ? 

7.  How  many  shot  are  there  in  a  rectangular  pile  of  twelve  courses 
if  the  uppermost  course  contains  .5  shot  ? 

8.  How  many  shot  are  there  in  a  triangular  pile  whose  lowermost 
course  contains  253  shot  ? 

9.  The  number  of  shot  in  a  certain  triangular  pile  is  four  sevenths  of 
the  number  in  a  square  pile  of  the  same  number  of  courses.  How  many 
shot  are  there  in  each  pile  ? 

10.  How  many  shot  are  there  in  a  rectangular  pile  whose  top  row 
contains  9  balls  and  whose  bottom  course  contains  240  balls? 

11.  Show  that  13  +  2'  +  •  •  •  +  71^  =  (1  +  2  +  •  •  ■  +  h)^. 

12.  Show  that  V  +  2*  +  ---  +  n*:^^^(n  +  1)  (2  n  +  1)  (3  n'^  +  3n-  1). 

13.  What  is  the  order  and  what  the  sum  of  the  first  n  terms  of  the 
progression  whose  ?ith  term  is  ji^  —  n  +  1  ?    n  {n  +  1)  {n  +  2)  / 6  ? 

14.  If  we  write  down  the  arithmetical  progressions  of  the  first  order 
in  which  d  =  1,  2,  3,  •  •  •  respectively  and  then  sum  each  progression  to 
one,  two,  three,  four,  •  •  •  terms,  we  obtain  the  following  sequences  of 
numbers,  called  respectively  the  triangular,  quadrangular,  pentagonal,  •  •  • 
numbers : 

1,  3,  6,   10,  •  •  •  ;  1,  4,  9,   16,  •  •  • ;  1,   5,   12,  22,  •  •  • ;  •  •  • . 
Show  that  in  the  A'th  of  tliese  se<]uences  the  7ith  term  and  the  sum 
of  the  first  n  terms  are  n{kn  -  k  +  2)/2  and   n{n  +  1)  {kn  -  A;  +  3)/0 
respectively. 

15.  Show  that  the  order  of  an  arithmetical  progression  of  any  order  is 
not  changed  by  adding  to  its  terms  the  corresponding  terms  of  an  arith- 
metical progression  of  a  lower  order. 

16.  Show  that  if  in  a  polynomial  of  the  nth  degree,  /(x),  we  substitute 
for  X  successive  terms  of  any  aiithmetical  progression  of  the  first  order, 
we  obtain  an  arithmetical  progression  of  the  nth  order;  and,  in  general, 
that  if  we  substitute  for  x  successive  terms  of  any  arithmetical  iirogression 
of  the  rth  order,  we  obtain  an  arithmetical  progression  of  the  nrth  order. 


METHOD   OF   DIFFERENCES  371 


INTERPOLATION 

Interpolation.  Suppose  that  y  is  known  to  depend  on  x  in  717 
such  a  manner  that  for  each  value  of  x  between  a  and  h,  y  has 
a  definite  value.  Suppose  also  that  the  values  of  y  which 
correspond  to  certain  of  these  values  of  x  are  actually  known. 
Then  from  these  known  values  it  is  possible,  by  a  process 
called  interpolation,  to  derive  values  of  y  corresponding  to 
other  values  of  x  between  a  and  b. 

This  process  is  employed  when  the  general  expression  for 
y  in  terms  of  x  is  unknown,  or  if  known  is  too  complicated 
to  be  conveniently  used  for  reckoning  out  particular  values 
of  y. 

Briefly  stated,  the  process  is  as  follows :  we  set  y  equal  to 
the  simplest  integral  expression  in  x  which  will  take  the  given 
values  and  then  derive  the  values  of  y  which  we  seek  from  this 
equation.  Of  course  the  values  thus  obtained  will  ordinarily 
be  only  approximately  correct. 

Method  of  undetermined  coeflBicients.     We  may  proceed  as  in     718 
the  following  example. 

Example.  For  x  =  2,  3,  4,  5  it  is  known  that  y  —  5,  4,  —  7,  —  34 ; 
find  y  when  x  =  5/2. 

Since  the  simplest  polynomial  in  x  which  will  take  given  values  for 
four  given  values  of  z  will  ordinarily  be  one  of  the  third  degree,  we 
assume  that 

y=:bo  +  hx  +  b.X^  +  63^', 

and  then  find  the  coefficients  60,  &i,  &2)  h  as  follows. 

5  =  60  +  2  61  +    4  62  +  8  63. 

4  =  60  +  361+    962+  2763. 

-  7  =  60  +  4  61  +  16  62  +  64  63. 

-  34  =  60  +  5  61  +  25  62  +  125  ftg- 
Solving  these  equations,  60  =  1,  61  =  —  2,  62  =  4,  63  =  —  1. 

Hence  y  =  I  —  2  x  +  i  x^  —  x^. 

Therefore,  when  x  =  5/2  we  have  y  =  1-5  +  25-  125/8  =  43/8. 


Since  y  =  5 

when  x  =  2, 

Since  y  =  4 

when  X  =  3, 

Since  y  =  -l 

when  X  =  4, 

Since  2/  =  -  34 

when  X  =  5, 

372  A   COLLEGE   ALGEBRA 

And  in  general,  if  r  +  1  values  of  ij  are  known,  say  the 
values  y  =  yu  y2,  ••■,  Vr+v  corresponding  to  a;  =  a-i,  Xg,  •••, 
cc^^i  respectively,  we  assume  that 

y  =  b^  +  l>,x  +  b^x^  +  ■■■  +  Kx%  (1) 

find  ht^,  bi,  ■■■,  b,.  by  the  method  just  illustrated,  and  then 
employ  (1)  as  a  formula  for  computing  y  for  values  of  x  inter- 
mediate to  Xi,  X2,  ■  ■  ■,  a^r  +  l- 

719         Method  of  differences.     When  x^,  Xo,---,  x.+i  are  consecutive 
integers,  the  formula  (1)  of  §  718,  may  be  reduced  to  the  form 

(X  —  CCi)  (X  —  Xo)    -,      , 

y  =  yr+ix-x,)ch+^ Y^ -d,  +  --- 

(x-x,)...(x-x,) 
^  1 . 2  •  •  •  r  '■'  ^  ^ 

where  di,  d^,  ■■-,  d,  denote  the  first  terms  of  the  successive 
orders  of  differences  of  ?/i,  y-i,---,  y,  +  i. 

For  since  Xi,  X2,  •  •  •,  a;,.  +  i  are  consecutive  integers,  the  corresponding 
values  oibo  +  biX-\ 1-  ^x*",  namely  yi,  2/2,  •  •  ■ ,  y,-  +  u  form  an  arithmet- 
ical progression  of  the  rth  order,  §  714.  Hence  we  may  also  obtain  2/1, 
2/2,  ••  •  by  substituting  n  =  1,  2,  •  •  •  in  the  formula,  §  710,  (I),  namely 

y  =  yi  +  {n-l)di  + ^-^ ch  +  ■■■  + i.2...r 

But  setting  n  =  1,  2,  3,  •  •  •  in  this  formula  will  give  identically  the 
same  results  as  setting  x  =  Xi,  X2,  X3,  •  •  •  =  Xi,  Xi  +  1,  Xi  -f-  2,  •  •  •  in  (2). 

Therefore  the  second  member  of  (2)  and  that  of  (1),  §  718,  have  equal 
values  for  r  +  1  values  of  x.  But  both  are  of  the  rth  degree.  Hence  they 
are  identically  equal,  §  421. 

Thus,  as  in  §  718,  for  x  =  2,  3,  4,  5,  let  y  =  5,  4,  -  7,  -  34.     We  have 

2/1,  2/2,  ^3,  2/4  =  5,  4,    -  7,   -  34. 
First  differences  -1,  -11,  -27. 

Second  differences  -  10,  -  10. 

Third  difference  -  6. 

Substituting  in  (2),  Xi  =  2,  Xa  =  3,  X3  =  4,  2/1  =  5,  di  =  -  1,  ''2  =  -  10, 
dg  =  -  6,  we  have  2/  =  5  -  (x  -  2)  -  5  (x  -  2)  (x  -  3)  -  (x  -  2) (x  -  3){x  -  4), 
which  may  be  reduced  to  2/  =  1  -  2  x  +  4  x^  _  x^,  as  in  §  718. 


METHOD   OF   DIFFERENCES  373 

Example.  Given  v^jjo  =  3.1072,  v'31  =  3.1414,  -v^^  3.1748,  and 
\/33  =  3.2075;  find  VsLC. 

yi,  2/2,   2/3,   2/4  =  3.1072,    3.1414,      3.1748,  3.2075. 
First  differences  .0342,       .0334,     .0327. 

Second  differences  -  .0008,   -  .0007. 

Third  difference  .0001. 

Substituting  in  (2)  Xi  =  30,  x^  =  31,  x^  =  32,  yi  =  3.1072,  di  =  .0342, 
da  =  -  .0008,  ds  =  .0001,  and  x  =  31.6,  we  have 

3.1072  +  (1.6)  (.0342)  +  11:M^(_  .0008)  +  (^-^H'^)  (~ '^^  (_  .0001) 

=  3.1072  +  .05472  -  .000384  +  .0000064  =  3.1615  +. 

Lagrange's  formula.     The  formula  (1)  of  §  718  may  also  be     720 
reduced  to  the  following  form,  due  to  Lagrange : 

_         (.r  —  a-o)  (x  —  Xs)  ■  ■  ■  (x  —  x,.^^) 
^  ~  ^'  (^1  -  x.^{xi  -  a-3)  •  •  ■  (X,  -  x,^,) 
(x-x,)(x-x,)---(x-x,.^,) 
^'  i^-,  -  ^i)(-^2  -  ^3)  •  •  •  (^2  -  *.  +  i) 

{x-x,){x-x^)---{x-x:) 

For  the  right  member  of  (3)  is  an  integral  function  of  x  of 
the  ?-th  degree  and  its  values  for  ic  =  a-^,  Xo,  ■■■,  x,.,^^  are  ?/i, 
Z/2)  ■■-,  l/r+i-  Thus,  if  we  set  x  =  x^,  every  term  except  the 
first  vanishes  and  the  first  term  reduces  to  yi.  Hence,  §  421, 
the  right  member  of  (3)  and  that  of  (1),  §  718,  are  equal  for 
r  +  1  values  of  x  and  are  therefore  equal  identically. 

Thus,  as  in  §  718,  for  x  =  2,  3,  4,  5,  let  2/  =  5,  4,  -  7,  -  34.  Sub- 
stituting in  (3),  we  obtain 

(x_3)_(x-_4)_(x-_5) 
(2  -  3)  (2  -  4)  (2  -  5) 
^(x-2)(x-4)(x-5)  _     (x-2)(x-3)(x-5)  _  ^^  (x-2)  (x-3)  (x-4) 
(3_2)(3-4)(3-5)         (4-2)  (4-3)  (4-5)      "    (5-2)  (5-3)  (5-4)' 
which  will  reduce  to  ?/  =  1  -  2  x  +  4  x^  -  x^,  as  in  §  718. 


374  A   COLLEGE    ALGEBRA 


EXERCISE  LXI 

1.  For  X  =  -  3,  -  2,  -  1,  0  it  is  known  that  ?/  =  -  20,  6,  0,  4 ;  find  y 
when  X  —  —  5/2,  also  when  x  =  —  1  /2. 

2.  Given  that  /(4)  =  10,  /(6)  =  -  12,  /(7)  =  -  20,  /(8)  =  -  18  ;  find 
/(x)  and  then  compute  /(12). 

3.  Giventhat252  =  625,  262  =  676,  272  =  729;  find  26.542  by  the  method 
of  differences. 

4.  Given  that  23  =  8,  3-3  =  27,  43  =  64,  53  =  125;  find  4.83  by  the 
method  of  differences. 

5.  Given  that  1/22  =  .04546,  1/23  =  .04348,  1/24  =  .04167,  and 
1/25  =  .04;  find  1/23.6  by  the  method  of  differences. 

6.  Given  that  V432  =  20.7846,  V433  =  20.8087,  V434  =  20.8327, 
V435  =  20.8566,  V436  =  20.8806  ;  find  V435.7  by  the  method  of  differ- 
ences. 

7.  By  aid  of  Lagrange's  formula  find  the  polynomial  of  the  third 
degree  whose  values  for  x  =  —  2,  0,  4,  5  are  5,  3,   —  2,  —  4. 


XXIV.     LOGARITHMS 

PRELIMINARY   THEOREMS    REGARDING  EXPONENTS 

721  Theorem  1.     If  a,  denote  any  real  numher  greater  than  1, 

p 
and  p,  q  denote  jmsitive  integers,  then  a^  >  1. 

For  a  >  1,  .-.  aP  >  1,  .-.  ^ai'  >  1,  .-.  a'/  >  1,  §  261. 

722  Theorem  2.     If  a,  denote  any  real  numher  greater  than  1, 
and  V,  s  any  two  rationals  such  that  r  >  s,  then  dJ  >  a^ 

For  r  -  s>0,  .-.  a''-''>l,  .-.  a'--'-  a'>a',  .-.  a'->a%  §§  721,  261. 

723  Theorem  3.     7/"a  >  1  and  n  be  integral,  then  ^^^"^  a"  =  00. 

Fur  since  a  >  1,  we  may  write  a  =  1  -\-  d,  where  d  is  positive. 

Then  a"  =  (1  +  d)",  and  since  (1 +  d)">l +  nd,  §  561,  wehave  a"  >l  +  nd. 

Therefore,  since  1™  (1  +  nd)  =  00,  we  have  1""  a"  =  00. 


LOGARITHMS  375 

Theorem  4.     If  0  <  a.  <  1,  and  n  be  integral,  lim  a"  =  0.  724 

For  let  a  =  1  /6,  where  6  >  1,  since  a  <  1. 

Then  li^"  a«  =  1  /J™  ^"  =  0,  §  512,  since  Jii^  6»  =  oo,  §  723. 

1 
Theorem  5.     If  u  ^»e  integral,  lii^i  V^  =  JiiJ  a"  =  1.  725 

1.  When  a  >  1,  we  have  a"  >  1,  §  721,  so  that  a"  =  1  +  d„,  where  d„  is 
some  positive  number  dependent  on  n. 

Tlien  a  =  (1  +  d«)»,  .-.  a>l  +  ?id«,  .-.  d„  <(a  -  \) / n. 
Tlierefore,  since  1™  (a  -  l)/n  =  0,  §  512,  we  have  Jij"^  cZ„  =  0. 

Hence  Ji|^  a"  =  jij"  (1  +  d„)  =  1. 

2.  When  0  <  a  <  1,  let  a  =  1/6,  where  &>1,  since  a<l. 

I  i  i 

Then  li"^  a"  =  1  /^l™  &«  =  1,  since  J™  6"  =  1,  by  1. 

Theorem  6.     If  h  he  a  rational  number  and  -x.  be  a  variable     726 
which  ajjjyroaches  b  through  rational  values,  then  ^^^^  a''  =  a^. 

1.  The  theorem  holds  true  when  6,  the  limit  of  x,  is  0. 

For  in  this  case  we  can  select  a  variable  n  which  takes  integral  values 
only  and  such  that  we  shall  always  have  —  l/n<x<l/n  and  that  when 
X  =  0  then  n  =  co.  i  j 

Then   a^  will   always    lie    between   a"   and    a  «,    §  722,    and   since 
i_  1 

Ihn  an  =  lim  a  «  =  1,  §  725,  we  have  li™  a='  =  l  =  a". 

2.  The  theorem  holds  true  when  6  ?^  0. 

For  since  a-*  =  a''  •  a^-'',  we  have  1*™^  a^  —  a^  •  ^^^  a^-*  =  a'',  by  1. 

Theorem  7.     If  h  be  an  irrational  number  aiid  y.  he  a  vari-     727 
able  which  approaches  b  through  rational  values,  then  a'^  will 
approach   a   limit   as  x  =  b,   and  the   value   of  this  limit  is 
independent  of  the  values  which  x  takes  i^i  approaching  b. 

The  reasoning  is  the  same  whether  a  >1  or  a  <  1,  but  to  fix  the  ideas 
we  shall  suppose  that  a  >  1. 

There  are  infinitely  many  sequences  of  rational  values  through  which 
X  may  run  in  approaching  h  as  limit.  From  among  them  select  some 
particular  increasing  sequence,  and  represent  x  by  x'  when  supposed  to 
run  through  this  sequence.  Then  as  x'  ==  b,  the  variable  a-^'  continually 
increases.  §  722,  but  remains  finite  —  less,  for  instance,  than  a"^,  if  c  denote 


376  A   COLLEGE    ALGEBRA 

any  rational  greater  than  b.     Hence  a-^'  approaches  a  limit,  §  192.     Call 
this  limit  L. 

It  only  remains  to  prove  that  a'  will  approach  this  same  limit  L  it  x 
approach  b  through  any  other  sequence  of  rationals  than  that  through  which 
x'  runs.  But  a^  =  a^  •  a^  -  ^'  and  therefore  lim  a""  -  lim  a^'  ■  lim  a*-  =='  =  L, 
since  lim  a^-^'  —  1,  §  726. 

728  Irrational  exponents.  We  employ  the  symbol  a^  to  denote 
the  limit  which  a^  will  approach  when  x  is  made  to  approach 
b  through  any  sequence  of  rational  values.  Hence  by  a'',  when 
b  is  irrational,  we  shall  mean  ^^]^^  a*. 

729  Having  thus  assigned  a  meaning  to  a^  when  x  is  irrational, 
we  can  readily  prove  that  li]^^  a""  =  a*  when  x  approaches  b 
through  a  sequence  of  irrational  values. 

For  let  x',  x,  x"  denote  variables  all  of  which  approach  b  as  limit,  x'  and 
x"  through  sequences  of  rational  values  and  x  through  a  sequence  of  irra- 
tional values,  and  such  that  x'  <x  <x".  It  then  follows  from  §§  726,  727 
that  a^  lies  between  a^'  and  a-^",  and  therefore  since  j,™  a^'  =  j,|2\  a^"  =  a*", 
that  1™  a^  =  a^. 

730  Theorem  8.  The  laivs  of  exponents  are  valid  for  irrational 
exp07ients. 

For  let  b  and  c  denote  irrational  numbers,  and  x  and  7j  variables  which 
approach  b  and  c  as  limits.  We  suppose  x  and  y  to  take  rational  values 
only. 

1.    a''-  a"  =  a''  +  '^. 

For  since  W^a"  =  a^  +  'J,  we  have  lim  a^a'J  =  lim  0^+". 


But 

lim  a^a'J  =  lim  a^  •  lim  av  =  a^a". 

§§  203,  728 

and 

lim  a^  +  y  =  a'""  (■>■+  s'>          -a^  +  <=. 

§§  203,  728 

2.  {a>'y  =  a'"'. 

For 

(a^)y  =  fF". 

Hence 

lim  (ax)w  -  jii"^  a'"",  or  (a}>)y  =  aPv. 

§728 

Hence 

lim  (aby  —  lini  (jb.-/^  or  {a!>Y  =  aK 

§§  728,  729 

3.  (aby  =  a'=¥. 

For 

(ab)?'  =  W'bv. 

Hence 

lim  (a/*)"  =  lim  a«by  =  lim  a"  •  lim  6v. 

§203 

That  is, 

(a6)'-  =  a'-^-. 

§728 

LOGARITHMS  377 


LOGARITHMS.     THEIR   GENERAL   PROPERTIES 

Logarithms.     Take  a,  any  positive  number  except  1,  as  a  base     731 
or  number  of  reference.     We  have  shown  that  every  real  power 
of  a,  as  a''^,  denotes  some  definite  positive  number,  as  m.     In  ,a 
subsequent  section  we  shall  show  conversely  that  every  positive 
number,  m,  may  be  expressed  in  the  form  a*^,  where  fi  is  real. 

If  a'^  =  m,  we  call  /a  the  loijaritlim  of  m.  to  the  base  a  and     732 
represent  it  by  the  symbol  log„?>i.     Hence  the  logarithm  of  m. 
to  the  base  a  is  the  exponent  of  the  power  to  which  a  must  be 
raised  to  equal  ?»,  that  is  a'""""'  =  m. 

Thus,  3*  =  81,  .-.  4  =  log381;  2-3  =  1/8,  .-.  -3  =  log.2l/8. 

Since  a"  =  1,  we  always  have  log„l  =  0;  and  since  a^  ■=  a,     733 
we  always  have  log,,  a  =  1. 

When  a  >  1,  it  follows  from  a^  =  m,  by  §  722,  that  to  any     734 
increase  in  the  number  m  there  corresponds  an  increase  in 
its  logarithm  p. ;  also  that  if  m  is  greater  than  1,  its  loga- 
rithm /A  is  positive,  and  that  if  m  lies  between  1  and  0,  its 
logarithm  /u,  is  negative. 

Again,  when  a  >  1,  we  have,  §  723,  735 

lim  a*^  =.  CO,  and  liii^  xr^  =  lim  1  /a'^  =  0. 
We  therefore  say,  when  a  >  1,  that  log„oo  =  co,  and  log„0  =  —  oo. 

Theorem  1.      The  logarithm  of  a  product  to  any  base  is  the     736 
su)n  of  the  logarithms  of  the  factors  to  the  same  base. 

For  let  m  =  a*^,  that  is  ^u  =  log„ ??;, 

and  n  —  ay^  that  is   v  =  log„ji. 

Then  9?in  =  a'^a'' =  a'-'  +  '', 

that  is,  loga»i?i=  ^  +  V  =  log,,  m  +  logaW. 

Theorem  2.      The  logarithm  of  a  quotient  is  the  logarithm  of     737 
the  dividend  minus  the  logarithm  of  the  divisor. 

For  if  m  =  a*^  and  n  =  a", 

we  have  rn/n  =  af^ /a""  =  a'^-'', 

that  is,  logam/n  =  fj.  —  v  =■  loga»i  —  log«ft. 


378  A   COLLEGE    ALGEBRA 

738  Theorem  3.  The  logarithm  of  any  power  of  a  number  is  the 
logarithm  of  the  number  multiplied  by  the  exponent  of  the 
power. 

For  if  m-a)^, 

we  have  nv  =  {af^y  =  a/^^, 

that  is,  loga  m''  =  r/j.  =  r  loga  m. 

739  Theorem  4.  The  logarithm  of  any  root  of  a  mimher  is  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 

For  if  m  =  a", 

we  have  V//i  =  Va*^  =  a% 

that  is,  loga  >/w  =  fi/s  =  (loga m)/s. 

740  The  practical  usefulness  of  logarithms  is  due  to  the  proper- 
ties established  in  §§  736-739.  Logarithms  of  numbers  to  the 
base  10  have  been  computed  and  arranged  in  tables.  If  we 
avail  ourselves  of  such  a  table,  we  can  find  the  value  of  a 
product  by  an  addition,  of  a  quotient  by  a  subtraction,  of 
a  power  by  a  multiplication,  and  of  a  root  by  a  division. 

Thus,  log— 1^  =  log  v^5  +  log  Ve  -  log  325  §§  736^  737 

=  (log  5) /7  +  (log  6) /8- 25  log  3.     §§738,739 

Hence,  to  obtain  the  value  of  v5  ViJ/O-s,  we  have  only  to  look  up  the 
values  of  log  5,  log  6,  and  log  3  given  in  the  table,  then  to  reckon  out 
the  value  of  (log  5)  /7  +  (log  0)  /8  -  25  log  3,  and  finally  to  look  up  in  the. 
table  the  number  of  which  this  value  is  tlie  logarithm. 

EXERCISE  LXII 

1.  Find  log24,  log42,  logv'gS,  log5025,  loga 729,  logio.OOl,  log2l/64, 
log2.125,  logaVa^,  log8l28,  log„2a3. 

2.  If  logio2  =  .3010  and  logio3  =  .4771,  find  the  logarithms  to  the 
base  10  of  12,  9/2,  v^,  Vg. 

3.  Express  loga 600^  in  terms  of  loga 2,  loga 3,  and  loga5. 


LOGARITHMS  379 

4.    Express  the  logarithms  of  each  of  the  following  expressions  to  the 
base  a  in  terms  of  loga&,  logaC,  logad. 

(1)  6^c~Vf^'-  (2)   ^/a-2v^**-^&Wa"^• 


■3V81'\^729-9-'  =  31/] 


5.    Prove  that  logs  V  81 V  729  •  9"  ^  =  31/18, 


X  4-  vx^TTT  

6.    Prove  that  log„  =  2  loga (x  +  Vx^  -  1). 


COMMON  LOGARITHMS 


Computation   of  common    logarithms.     For    the   purposes    of     741 
numerical  reckoning  we  employ  logarithms  to  the  base  10. 
These  are  called,  common  logarithms.     In  what  follows  log  m 
will  mean  logi„?». 

We  have  10°  =  1,   .-.  log  1  =  0 ;   10^  =  10,  .-.  log  10  =  1 ;     742 
102  ^  100,  .-.  log  100  =  2,  ■  •  • ;  also  10"'  =  .1,  .-.  log  .1  =  -  1 ; 
10-2  =  .01,  .-.log  .01  =-2,  •••. 

Hence,  for  the  numbers  whose  common  logarithms  are  iyite- 
gers  we  have  the  table  : 


The  numbers     • 

•  •  .001, 

.01, 

.1,      1, 

10, 

100, 

1000,  • 

eir  logarithms  • 

•  -3, 

_  9 

-1,    0, 

1, 

2, 

3,- 

Observe  that  in  this  table  the  numbers  constitute  a  geomet- 
rical progression  in  which  the  common  ratio  is  10,  and  the 
logarithms  an  arithmetical  progression  in  which  the  common 
difference  is  1. 

The  numbers  in  this  table  are  the  only  rationals  whose  743 
common  logarithms  are  rational,  for  all  fractional  powers  of 
10  are  irrational.  But,  as  we  proceed  to  show,  every  positive 
number  has  a  common  logarithm,  and  the  value  of  this  loga- 
rithm may  be  obtained  correct  to  as  many  places  of  decimals 
as  may  be  desired. 

If  we  extract  the  square  root  of  10,  the  square  root  of  the 
result  thus  obtained,  and  so  on,  continuing  the  reckoning  in 


380  A    COLLEGE    ALGEBRA 

each  case  to  the  fifth  decimal  figure,  we  obtain  the- following 
Jable : 

10^=3.16228,  10^^=1.07461,  10"'  =  1.00451, 

10^  =  1.77828,  W'  =  1.03663,  10"^*  =  1.00225, 

10^  =  1.33352,  10"«  =  1.01815,  10=^"  =  1.00112, 

10^"*  =  1.15478,  10"«  =  1.00904,  10"'"  =  1.00056, 

and  so  on,  the  results  obtained  approaching  1  as  limit  as 
we  proceed  (compare  §  725).  The  exponents  1/2,  1/4,  •  •  •  on 
the  left  are  the  logarithms  of  the  corresponding  numbers  on 
the  right. 
744  By  aid  of  this  table  we  may  compute  the  common  loga- 
rithm of  any  number  between  1  and  10  as  in  the  following 
example. 

Example.     Find  the  common  logarithm  of  4.26. 

Divide  4.26  by  the  next  smaller  number  in  the  table,  .3.16228. 

The  quotient  is  1.34719.     Hence  4.26  =  3.16228  x  1.34719. 

Divide  1.34719  by  the  next  smaller  number  in  the  table,  1.33352. 

The  quotient  is  1.0102.     Hence  4.26  =  3.16228  x  1.33352  x  1.0102. 

Continue  thus,  always  dividing  the  quotient  last  obtained  by  the  next 
smaller  number  in  the  table. 

If  q,,  denote  the  quotient  in  the  nth  division,  we  shall  obtain  by  this 
method  an  expression  for  4.26  in  the  form  of  a  product  of  n  numbers 
taken  from  the  table  and  q„,  the  result  being 

4.26  =  3.16228  x  1.33352  x  1.00904  x  •  •  •  x  g„ 

=  10*  •  10^  .  lO'^s . .  •  g„  =  10^  +  s  +  =^= '°  " '"'"'  g„. 

As  n  increases,  the  exponent  1/2  +  1/8  +  1/256  +  •  •  •  to  n  terms  also 
increases.  But  it  remains  less  than  1,  since  it  is  always  a  part  of  the  infi- 
nite series  1/2  +  1/4  +  1/8  +  •  •  •  who.se  sum  is  1,  §  704,  Ex.  1.  Hence 
it  approaches  a  limit  which  is  some  number  less  than  1,  §  192.  Represent 
this  limit  thus  :   1/2  +  1/8+1  /256  +  •  •  • . 

Again,  as  n  increa.ses,  q„  approaches  1  .as  limit.  For  each  quotient 
lies  between  the  divisor  used  in  obtaining  it  and  1,  and  as  the  process  is 
continued  the  divi.sors  approach  1  as  limit. 

Hence 4.26  =  1*/"  10=  +  s  +  ^Js  +  •  •  • ""' '""■»(,„  =  10^  +k  +  '.h  +  ---^  and  there- 
fore log 4.26  =  I'/V+l/B  +  1/256  +  •  •  •  =  .6294  •  •  •• 


LOGARITHMS  381 

From  the  common  logarithms  of  the  numbers  between  1  and     745 
10  we  may  derive  the  common  logarithms  of  all  other  positive 
numbers  by  the  addition  of  positive  or  negative  integers. 

Example.     Find  the  common  logarithms  of  42.6  and  .426. 

1.  We  have  42.6  =  10  x  4.26. 
Hence                 log  42.6  =  log  10  +  log  4.26 

=  1  +  log  4.26  =  1.6294. 

2.  Again,  .426  =  4.26/10  =  10-1x4.26. 
Hence                 log  .426  =  log  lO-i  +  log  4.26 

=  -1  + log  4. 26  =  1.6294. 

Similarly  log  426  =  2.6294,  log  .0426  =  2.6294,  and  so  on ;  that  is,  we 
may  obtain  the  logarithm  of  any  number  which  has  the  same  sequence  of 
figures  as  4.26  by  adding  a  positive  or  negative  integer  to  log  4.26. 

Characteristic  and  mantissa.     In  a  logarithm  expressed  as  in     746 
the  preceding  example  we  call  the  decimal  part  the  mantissa 
and  the  integral  part  the  characteristic. 

Thus,  log  42.6  =  1.6294  and  log  .426  =1.6294  have  the  same  mantissa 
.6294  and  the  characteristics  1  and  —  1  respectively. 

As  has  been  shown  in  §  745,  if  n  denote  a  positive  number     747 
expressed    as    an    integer   or  decimal,   the  inaiitissa   of  log  n 
depends  solehj  on  the  sequence  of  figures  in  n,  a7id  the  charac- 
teristic on  the  2^08  it  ion  of  the  decimal  point  in  n. 

If  n  lies  between  1  and  10,  that  is,  if  it  has  but  one  figure  748 
in  its  integral  part,  the  characteristic  of  log  7i  is  0,  §  744.  If 
we  shift  the  decimal  point  in  such  a  number  n  one  place  to 
the  right,  that  is,  if  we  multiply  7i  by  10,  we  add  1  to  the 
0  characteristic  of  log  n.  Similarly  if  we  shift  the  decimal 
point  two  places  to  the  right,  we  add  2  to  the  characteristic  of 
log  n,  and  so  on,  §  745.     Hence  the  rule  : 

If  n>l,  the  characteristic  of  log  n  is  one  less  than  the  member 
offgures  in  the  integral  part  of  n. 

Thus,  log  426000  =  5.6294,  log  42600000  =  7.6294,  and  so  on. 


382  A    COLLEGE    ALGEBRA 

749  In  like  manner,  if  in  a  number  n  which  has  but  one  figure 
in  its  integral  part  we  shift  the  decimal  point  yx  places  to  the 
left,  that  is,  if  we  multiply  n  by  10"'^,  we  add  —  /a  to  the  0  char- 
acteristic of  log  n.  Thus,  log  .426  =  -  1  +  log  4.26  =  1.6294, 
log  .0426  =  2.6294,  and  so  on,  §  745. 

In  practice  we  find  it  convenient  to  write  these  negative 
characteristics  1,  2,  •  •  ■  in  the  form  9  —  10,  8  —  10,  •  •  •,  and  to 
place  the  positive  part  9,  8,  •  ■  •  before  the  mantissa,  and  the 
-  10  after  it.  Thus,  instead  of  1.6294  we  write  9.6294  -  10. 
Hence  the  rule : 

Ifn<l,the  characteristic  of  log  n  is  negative.  To  obtain  it, 
subtract  from  9  tJie  7iumber  of  O's  immediately  to  the  rigid  of  the 
decimal  jioint  in  n,  then  write  the  result  before  the  mantissa, 
and  —  10  after  it. 

Thus,  log  .00426  =  7.6294  -  10,  log  .000000426  =  3.6294  -  10. 

If  more  than  nine  but  less  than  nineteen  O's  immediately  follow  the 
decimal  point,  subtract  their  number  from  19  and  write  the  result  before 
the  mantissa  and  —  20  after  it ;  and  so  on. 

Example.     Given  log  2  =  .3010,  find  the  number  of  digits  in  2^5. 

750  A  table  of  logarithms.  The  accompanying  table,  pp.  384,  385, 
contains  the  mantissas  of  the  logarithms  of  all  numbers  of 
three  figures  computed  to  the  fourth  place  of  decimals  and 
arranged  in  rows  in  the  order  of  their  magnitude,  the  decimal 
points  before  the  mantissas  being  omitted. 

From  this  table  we  may  also  derive  mantissas  for  numbers 
of  more  than  three  figures  by  aid  of  the  principle : 

When  a  number  is  changed  by  an  amount  which  is  very  small 
in  comparison  tvit?i  the  number  itself,  the  change  in  the  loga- 
rithm of  the  number  is  nearly  proportional  to  the  change  in  the 
number. 

Numerical  results  obtained  by  aid  of  this  table  are  not  to 
be  trusted  beyond  the  fourth  figure.  When  greater  accuracy 
is  required  we  must  use  tables  in  which  the  mantissas  are 


LOGARITHMS  383 

given  to  more  than  four  places  of  decimals.     The  student  will 
find  it  easy  to  procure  a  five-,  six-,  or  seven-place  table. 

To  find  the  logarithm  of  a  number  from  this  table.  We  proceed 
as  in  the  following  examples.  , 

Example  1.     Find  the  logarithm  of  .00589. 

We  look  up  the  first  two  significant  figures,  58,  in  the  column  headed 
N  in  the  table,  then  run  along  the  row  to  the  right  of  58  until  the  column 
is  reached  which  is  headed  by  the  third  figure,  9.  We  there  find  7701. 
This  (with  a  decimal  point  before  it)  is  the  mantissa  sought.  The  char- 
acteristic is  7  —  10,  §  749.     Hence 

log  .00589  =  7.7701 -10. 

Example  2.     Find  the  logarithms  of  8  and  46. 

The  mantissas  of  these  logarithms  are  the  same  as  those  of  800  and 
460  respectively.     Hence,  proceeding  as  in  Ex.  1,  we  find 
log  8  =  .9031,  log46  =  1.6628. 

Example  3.     Find  the  logarithm  of.  4673. 

The  mantissa  is  the  same  as  that  of  log  467.3.  It  must  therefore  lie 
between  mant.  log  467  and  mant.  log  468. 

From  the  table  we  find  mant.  log  467  =  6693  and  mant.  log  468  =  6702, 
and  the  difference  between  these  mantissas  is  9. 

,     Thus  if  we  add  1  to  467,  we  add  9  to  mant.  log  467.     Hence  if  we  add 
i.3  of  1  to  467,  we  should  add  .3  of  9,  or  3  approximately,  to  mant.  log  467.  . 
'     Hence  mant.  log  467.3  =  6693  -|-  3  =  6696,  and  therefore 
log  .4673  =  9.6696  -  10. 

Observe  that  until  the  characteristic  is  introduced  we  omit 
the  decimal  point  which  properly  belongs  before  the  mantissa. 

The  method  illustrated  in  Ex.  3  for  finding  the  mantissa  of 
the  logarithm  of  a  number  of  more  than  three  figures  may  be 
described  as  follows  : 

From  the  table  obtain  m,  the  mantissa  corresponding  to  the 
first  three  figures,  also  d,  the  difference  between  m  and  the  next 
greater  mantissa. 

Multiply  d  by  the  remaining  part  of  the  number  with  a 
decimal  point  before  it,  aiid  add  the  integral  part  of  the  product 
{increased  by  1  if  the  decimal  part  is  .5  or  more)  to  m. 


384 


A    COLLEGE    ALGEBRA 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0765 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2766 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4466 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

6024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

6172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

6289 

5302 

34 

5315 

5328 

5340 

5353 

6366 

5378 

5391 

5403 

6416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5()58 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5776 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6326 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6622 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

mr>r, 

()(■)(  )5 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

()7r]S 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6S.",9 

6S4K 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

62 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

LOGARITHMS 


385 


N 

0 

1 

a 

3 

^ 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

755!) 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7684 

7642 

7649 

7667 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8625 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8003 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9730 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

386  A    COLLEGE   ALGEBRA 

752  To  find  a   number  when  its  logarithm  is  given.     We  have 

merely   to    reverse    the   process   described    in   the   preceding 
section. 

Example  1.     Find  the  number  whose  hjgarithm  is  5.9552  —  10. 

We  find  the  mantissa  9552  in  the  table  in  the  row  marked  90  and  iu 
the  column  marked  2.     Hence  the  required  sequence  of  figures  i.s  902. 

But  since  the  characteristic  is  5  —  10,  the  number  is  a  decimal  with 
9  —  5,  or  4,  O's  at  the  right  of  the  decimal  point,  §  749.  Hence  the  required 
number  is  .0000902. 

Example  2.     Find  the  number  whose  logarithm  is  7.5520. 

Looking  in  the  table  we  find  that  the  given  mantissa  5520  lies  between 
the  mantissas  5514  and  5527  corresponding  to  356  and  357  respectively; 
The  lesser  of  these  mantissas,  5514,  differs  from  the  greater,  5527,  by  13 
and  from  the  given  mantissa,  5520,  by  6. 

Thus,  if  we  add  13  to  the  mantissa  5514,  we  add  1  to  the  number  356, 
Hence  if  we  add  6  to  the  mantissa  5514,  we  should  add  6/13  of  1,  or  .$ 
approximately,  to  356. 

Hence  the  required  sequence  of  figures  is  3565,  and  therefore  by  the 
rule  for  characteristic,  §  748,  the  required  number  is  35650000. 

We  therefore  have  the  following  rule  for  finding  the  sequence 
of  figures  corresponding  to  a  given  mantissa  which  is  not  in 
the  table : 

Fi7id  front  the  table  the  next  lesser  mantissa  m,  the  three 
corresponding  figures,  and  d,  the  difference  between  m  and  the 
next  greater  mantissa.  ; 

Subtract  va.  from  the  given  mantissa  and  divide  the  remainder 
by  d,  an7iexi7ig  the  resulting  figure  to  the  three  figures  already 
obtained. 

753  Cologarithms.     The  cologarithm.  of  a  number  is  the  logarithm 
of  the  reciprocal  of  the  number. 

Since  colog  m  =  log  l/m=z  log  l  —  \ogm  =  —  log  m,  §§  733, 
737,  we  can  find  the  cologarithm  of  a  number  by  merely 
changing  the  sign  of  its  logarithm.  But  to  avail  ourselves  of 
the  table  we  must  keep  the  decimal  parts  of  all  logarithms 
positive.     We  therefore  proceed  as  follows : 


LOGARITHMS  387- 

Example  1,  Find  colog  89.2. 

We  have  log  1  =  10          — 10 

and  log  89. 2=    1.9504 

Hence  colog  89.2=    8.0496-10 

Example  2.  Find  colog. 929. 

We  have  log  1  =  10           - 10 

and  log  .929  =    9.9680  -  10 

Hence  colog.  929=     .0320 

Hence  we  may  find  the  cologarithm  of  a  number  from  its 
logarithm  by  beginning  at  the  characteristic  and  subtracting 
each  figure  from  9  until  the  last  significant  figure  is  reached, 
which  figure  must  be  subtracted  from  10.  To  this  result  we 
do  or  do  not  afiix  —  10  according  as  —  10  is  not  or  is  affixed 
to  the  logarithm.  In  this  way  when  the  number  has  not  more 
than  three  figures  we  may  obtain  its  cologarithm  directly  from 
the  table. 

Computation  by  logarithms.     The  following   examples  will     754 
serve  to  show  how  expeditiously  approximate  values  of  prod- 
ucts, quotients,  powers,  and  roots  of  numbers  may  be  obtained 
by  aid  of  logarithms  (compare  §  740). 

Example  1.     Find  the  value  of  .0325  x  .0425  x  5.26. 

Log  (.0325  X  .6425  x  5.26)  =  log  .0325  +  log  .6425  +  log  5.26. 

But  log. 0325=    8.5119-10 

log  .6425=    9.8079-10 

log    5.26  =      .7210 
Hence  log  of  product  =  19.0408  -  20  =  9.0408  -  10 

Therefore  the  product  is  .1099. 

Example  2.     Find  the  value  of  46. 72/. 0998. 
Log  (46. 72/. 0998)  =  log  46.72  -  log.  0998. 
But  log  46.72  =  11.6695  -  10 

log  .0998  =    8.9991  -  10 
Hence  log  of  quotient  =    2.6704 

Therefore  the  quotient  is  468.2. 


388  A   COLLEGE   ALGEBRA 

We  write  log  46.72,  that  is  1.6G95,  in  the  form  11.6695  -  10  in  order 
to  make  its  positive  part  greater  than  that  of  8.9991  —  10  which  is  to  be 
subtracted  from  it. 

Example  3.     Find  the  value  of  295  x  .05631  -^  806. 

Log  (295  X  .05631  --  806)  =  log  295  +  log  .05631  +  colog  806. 

But  log  295=    2.4698 

log  .05631  =    8.7506-10 
colog  806=    7.0937  -10 
Hence  log  of  required  result     =  18.3141  -  20  =  8.3141  -  10 
Therefore  the  required  result  is  .02061. 
Example  4.     Find  the  sixth  power  of  .7929. 
Log  (.7929)6  =  6  X  log  .7929. 
But  log  .7929  =    9.8992  -  10 


Hence  log  (.7929)6  =  59.3952  _  60  =  9.3952  -  10 

Therefore  (.7929)6  ^  .2484. 

Example  5.     Find  the  seventh  root  of  .( 


Log  V.00898  =  (log  .00898)  -4-  7. 

But  log  .00898  =    7.9533  -  10 

7)67.9533-70 

Hence  log  V.00898  =    9.7076-10 


Therefore  v.00898  =  .510. 

Observe  that  when  as  here  we  have  occasion  to  divide  a  negative  loga- 
rithm by  some  number,  we  add  to  its  positive  and  negative  parts  such  a 
multiple  of  10  that  the  (luotient  of  the  negative  part  will  be  —  10. 

Negative  numbers  do  not  have  real  logarithms  to  the  base 
10  since  all  real  powers  of  10  are  positive  numbers.  If  asked 
to  find  the  value  of  an  expression  which  involves  negative 
factors,  we  may  first  find  the  absolute  value  of  the  expression 
by  logarithms  and  then  attach  the  appropriate  sign  to  the 
result. 

Thus,  if  the  given  expression  were  456  x  (  -  85.96),  we  should  first  find 
the  value  of  450  x  85.96  by  logarithms  and  then  attach  the  —  sign  to  the 
result. 


LOGARITHMS 

EXERCISE   LXni 

Find  approximate  values  of  tlie  following  by  aid  of  logarithms. 

1.    79  X  470  X  .982. 

2.   (-  9503)  X  (-  .0086578). 

3.    1375600  X  8799000. 

4.    .0356  X  (-  .00049). 

n     8075 

^-    364.9'                          ^• 

.00542                                       24617 
.04708                                    -  .00054 

.643  X  7095 
■    67  X  9  X  .462 

9097  X  5.4086 
-  225  X  593  X  .8665 

10.   (2.388)5.                    11. 

(.57)-".                         12.   (19/11)9. 

13.   (1.014)26.                   14. 

V67.54.                         15.    V- .30892. 

16.    8l                             17. 

(.001)3.                           18.    (29t\)'. 

19.    V|  X  -v/ll.              20. 

VjL- (.009)1              21.    (.00068)-^ 

22.   (6|)3-'.                       23. 

(-9306)'.                     24.    (.0057)2-5. 

25.   (5648)^  X  (-  .94)3. 

26.    289273  -  (.8)1 

V.0476  X  V222 

„„      V943  X  7298 

^5059  X  .0088 

V.00006  X  .99 

^g         /     854  X  V.042 
\  7.9856  X  V.0005 

3Q       3/7*x92^x(.01)J 
\  (.00026)5  X  5968i 

389 


SOME  APPLICATIONS   OF   COMMON   LOGARITHMS 

Logarithms  to  other  bases  than  10.     From  the  logarithm  of  a     755 
number  to  the  base  10  we  can  derive  its  logarithm  to  any- 
other  positive  base  except  1  by  aid  of  the  theorem  : 

The  logarithms  of  a  number  m  to  two  different  bases,  a  and  b, 
are  connected  by  the  formula  log,,  m  =  log^  m/log^  b. 

For  let  7/1  =  a**,   that   is  /i  =  log„  7?i, 

and  let  b  =  a'',    that   is    ;/  =  log„6. 

1 
Since  a"  =  b,  we  have   a  =  b^. 

1  M. 

Hence  7?i  =  a*^  =  (6'')'^  =  6", 

that  is,  logi,  m  —  fx/v  —  loga  m  /  loga  6. 


390  A    COLLEGE   ALGEBRA 

Example.     Find  the  logarithm  of  .586  to  the  base  7. 
Tna     ^«fi      logio  .586  _  9.7679- 10  _      2321  _  _ 

We  reduce  9.7679  -  10  to  the  form  of  a  single  negative  number,  namely 
—  .2.321,  and  perform  the  final  division  by  logarithms. 

756  When  m  =  a  the  formula  gives  log^a  =  l/log^b. 

757  The  only  base  besides  10  of  which  any  actual  use  is  made 
is  a  certain  irrational  number  denoted  by  the  letter  e  whose 
approximate  value  is  2.718.  Logarithms  to  this  base  are 
called  natural  logarithms.  We  shall  consider  the;m  in  another 
connection. 

758  Exponential  and  logarithmic  equations.  Equations  in  which 
the  unknown  letter  occurs  in  an  exponent  or  in  a  logarithmic 
expression  may  sometimes  be  solved  as  follows. 

Example  1.     Solve  the  equation  IS^^+s  =  14^+7. 

Taking  logarithms  of  both  members,  (2  x  +  5)  log  13  =  (x  +  7)  log  14. 

o  ,  .  7  log  14  -  5  log  13      2.4532      „  _„ 

Solving,  X  = 2 5 —  = =  2.268. 

^'  2  log  13  -  log  14        1.0817 


Example  2.     Solve  the  equation  log  Vx  —  21  +  |  log  x  =  1. 
By  §§  736,  739,  we  can  reduce  this  equation  to  the  form 

log  Vx(x-21)  =  1  =  log  10. 
Hence  x2  _  21  x  =  100. 

Solving,  X  =  25  or  —  4. 

Example  3.     Solve  the  equation  x-^°s^  =  10 x. 
Taking  logarithms,       2  (log  x)2  =  log  x  +  1. 
Solving  for  log X,  logx  =  l  or  —1/2. 

Hence  x  =  10  or  1/VlO. 

759  Compound  interest.  Suppose  that  a  sum  of  P  dollars  is  put 
at  compound  interest  for  a  period  of  n  years,  interest  being 
compounded  annually  and  the  interest  on  one  dollar  for  one 
year  being  r. 

Then  the  amount   at  the   end   of   the   first   year  will  be 
P  +  Pr  or  P  (1  +  r),  at  the  end  of  the  second  year  it  will 


LOGARITHMS  391 

be  P(i  +  r)  •  (1  +  r)  or  P(l  +  ?•)^  and  so  on.     Hence,  if  A 
denote  the  amount  at  the  end  of  the  nth  year,  we  have 

I  A^P(l-\-ry. 

If  interest  be  compounded  semiannually,  A  =  P  (1  -{-  r/2)^"; 
if  quarterly,  A  =  P (I  +  r /AY" ;  and  so  on. 

We  call  P  the 2irese7it  worth  of  A.  If  A,  n,  and  r  be  given, 
we  can  find  P  by  means  of  the  formula  P  =  A(\  +  r)'". 

Example  1.  Find  the  amount  of  §2500  in  eighteen  years  at  4%  com- 
pound interest. 

We  have  log  A  =  log  2500  +  18  log  1.04  =  3.7039. 

Hence  A  =  $5057,  approximately. 

Example  2.  At  the  beginning  of  each  of  ten  successive  years  a  premium 
of  $120  is  paid  on  a  certain  insurance  policy.  What  is  the  worth  of  the 
sum  of  these  premiums  at  the  end  of  the  tenth  year  if  computed  at  4% 
compound  interest? 

The  required  value  is  120 [1.04  +(1.04)2  +  . ..  ^(i.o4)io], 

that  is,  by  §  701,  120  x  1.04  x  ^^ 

1.04  —  1 

By  logarithms,  (1.04)1"  -  1.479. 

Hence  the  required  value  is  120  x  1.04  x  .479  h-  .04;  and  this,  com- 
puted by  logarithms,  gives  $1494,  approximately. 

Annuities.     A  sum  of  money  which  is  to  be  paid  at  fixed     760 
intervals,  as  annually,  is  called  an  annuity. 

It  is  required  to  find  the  present  worth  of  an  annuity  of  A 
dollars  payable  annually  for  n  years,  beginning  a  year  hence, 
the  interest  on  one  dollar  for  one  year  being  r. 

The  present  worth  of  the  first  payment  is  ^  (1  +  r)~S  that 
of  the  second  payment  is  A  (1  +  r)"^,  and  so  on. 

Hence  the  present  worth  of  the  whole  is,  §  701, 

If  the  annuity  hejierpetual,  that  is,  if  n  =  oo,  then  (1  +  r)"  =  oo, 
and  the  formula  for  the  present  worth  reduces  to  A  /r. 


392  A  COLLEGE    ALGEBRA 

Example.  What  sum  should  be  paid  for  an  annuity  of  $1000  payable 
annually  for  twenty  years,  money  being  supposed  to  be  worth  3%  per 
annum  ? 

The  present  worth,  P,  is 1 • 

.03    L         (L03)2oJ 

By  logarithms,  we  find  that  (1.03)2o  =  1.803. 

„  „      1000  r,  1    1       1000  X  .803      ^,,„,, 

Hence  P  = 1 = =  $14845,  approximately. 

.03    L        1.803J       .03  X  1.803  ,    h-f         ^       J 


EXERCISE  LXIV 

1.  Find  logs 555,  logy. 0463,  logioo47. 

2.  Solve  the  following  exponential  equations. 

(1)3^  =  729.  (2)  a^=  +  2  =  a3^.  (3)213^  =  516-^+4. 

3.  Solve  the  following  logarithmic  equations. 

(1)  log  X  +  log  (X  +  3)  =  1.  (2)  log  x2  +  log  X  =  2. 

(3)  log  (1-2  x)3  -  log  (3  -  x)3  =  6.     (4)  x'^g^  =  2. 

4.  Find  the  amount  of  $7500  in  thirty-five  years  at  5%  compound 
interest,  the  interest  being  compounded  annually. 

5.  Find  the  amount  of  $5500  in  twenty  years  at  3%  compound  interest, 
the  interest  being  compounded  semiannually. 

6.  Show  that  a  sum  of  money  will  more  than  double  itself  in  fifteen 
years  and  that  it  will  increase  more  than  a  hundredfold  in  ninety-five 
years  at  5%  compound  interest. 

7.  What  sum  will  amount  to  $1250  if  put  at  compound  interest  at  4% 
for  fifteen  years? 

8.  A  man  invests  $200  a  year  in  a  savings  bank  which  pays  S},%  per 
annum  on  all  deposits.  What  will  be  the  total  amount  due  him  at  the 
end  of  twenty-five  years  ? 

9.  What  sum  should  be  paid  for  an  annuity  of  $1200  a  year  to  be 
paid  for  thirty  years,  money  being  supposed  to  be  worth  4%  per  annum  ? 
What  sum  should  be  paid  were  this  annuity  to  be  perpetual  ? 

10.  If  c  denote  the  length  of  the  hypotenuse  of  a  right-angled  triangle 
and  a,  b  the  lengths  of  the  other  two  sides,  b  =  V(c  +  a)(c  —  a).  Given 
c  =  586.4,  a  =  312.2,  find  b  and  the  area  of  the  triangle,  using  logarithms. 


TERMUTATIOXS   AND    COMBIXATIONS  393 

11.  If  ft,  6,  c  denote  the  lengths  of  the  sides  of  a  triangle  and 
s  =  {a  +  b  +  c)/2,  the  area  of  the  triangle  is  \''s{s  —  a){s  —  b)  {s  —  c). 
Find  the  area  of  the  triangle  in  which  a  =  410.8,  b  —  424,  c  =  25.68. 

12.  Find  the  area  of  the  surface  and  the  volume  of  a  sphere  the  length 
of  whose  radius  is  23.6  by  aid  of  the  formulas  S  —  4i7tr'\  F=4  7rr*/3, 
assuming  that  jr  =  3.1416. 


XXV.  PERMUTATIONS  AND  COMBINATIONS 

Definitions  of  permutation  and  combination.     Suppose  a  group     761 
of  n  letters  to  be  given,  as  a,  b,  c,  ■■■,  k,  denoting  objects  of 
any  kind. 

Any  set  of  r  of  these  letters,  considered  without  regard  to 
order,  is  called  a  combination  of  the  n  letters  t  at  a  time,  or, 
more  briefly,  an  v-covibinution  of  the  n  letters. 

We  shall  use  the  symbol  C"  to  denote  the  number  of  such 
combinations. 

Thus,  the  2-combinations  of  the  four  letters  a,  &,  e,  d  are 

ab,  ac,  ad,  be,  bd,  cd. 
There  are  six  of  these  combinations,  that  is,  Ct  =  6. 

On  the  other  hand,  any  arrangement  of  r  of  these  n  letters 
in  a  definite  order  in  a  row  is  called  a  permutation  of  the  n 
letters,  r  at  a  time,  or,  more  briefly,  an  T-perynutation  of  the 
71  letters. 

We  shall  use  the  symbol  P".  to  denote  the  number  of  such 
permutations. 

Thus,  the  2-permutations  of  the  four  letters  a,  b,  c,  d  are 
ab,  ac,  ad,  be,  bd,  cd, 
ba,  ca,  da,  cb,  db,  dc.  ^ 

There  are  twelve  of  these  permutations,  that  is  P^  =  12.      ~^ 
Observe  that  while  ab  and  ba  denote  the  same  combination,  they  denote 
different  permutations. 

In  what  has  just  been  said  it  is  assumed  that  the  letters 
a.  b,  ■  ■  ■,  k  are  all  different  and  that  the  repetition  of  a  letter 


394  A    COLLEGE    ALGEBRA 

within  a  permutation  or  combination  is  not  allowed.     This 
will  be  the  understanding  throughout  the  chapter  except  where 
the  contrary  is  stated. 
763         A  preliminary  theorem.     AVe  have  already  had  occasion  to 
apply  the  following  principle,  §  554  : 

If  a  certain  thing  can  he  done  in  m  waijs,  and  if,  vlien  it  has 
been  done,  a  certain  other  thing  can  he  done  in  n  loays,  the  entire 
number  of  ways  in  which  both  things  can  be  done  in  the  order 
is  mn. 


We  reason  thus  :  Since  for  each  way  of  doing  the  first  thing 
there  are  n  ways  of  doing  both  things,  for  m  ways  of  doing  the 
first  thing  there  are  mn  ways  of  doing  both  things. 

More  generally,  if  a  first  thing  can  be  done  in  m  ways,  then 
a  second  thing  in  n  ways,  then  a  third  in  p  ways,  and  so  on, 
the  entire  number  of  ways  in  which  all  the  things  can  be  done 
in  the  order  stated  is  7n  -n-jy  ■  •■• 

Example.  How  many  numbers  of  three  different  figures  each  can  be 
formed  with  the  digits  1,  2,  3,  •  •  •,  9  ? 

We  may  choose  any  one  of  the  nine  digits  for  the  first  figure  of  the 
number,  then  any  one  of  the  remaining  eight  digits  for  its  second  figure, 
and  finally  any  one  of  the  seven  digits  still  remaining  for  its  third  figure. 
Kence  we  may  form  9  ■  8  •  7,  or  504,  numbers  of  the  kind  required. 

763  The  number  of  r-permutations  of  n  different  letters.  By  the 
reasoning  employed  in  the  preceding  example  we  readily  prove 
that  this  number  P'^  is  given  by  the  formula 

P'^  —  n(n  —  1)  (n  —  2)  ■•■  to  r  factors.  (1) 

For  in  forming  an  ?'-permutation  of  n  letters  we  may  choose 
any  one  of  the  n  letters  for  its  first  letter,  then  any  one  of  the 
remaining  n  —  1  letters  for  its  second  letter,  then  any  one  of 
the  n  —  2  letters  still  remaining  for  its  third  letter,  and  so  on. 

Hence,  §  762,  the  entire  number  of  ways  in  which  we  may 
choose  its  first,  second,  third,  ■  •  • ,  rth  letters,  in  other  word? 


PERMUTATIONS    AND   COMBINATIONS  395 

the  entire  number  of  ways  in  which  we  may  form  an  /--permu- 
tation with  the  n  letters,  is  n  («  —  1)  (w  —  2)  •  •  •  to  r  factors. 

Thus,  the  numbers  of  permutations  of  t^^  letters  a,  6,  c,  d,  e  one,  two, 
three,  four,  five  at  a  time  are 
P\  =  b,  P|  =  5  •  4,  P-]  =  5  ■  4  •  3,  P]  =  5  •  4  •  3  •  2,  P^  =  5  •  4  •  3  •  2  •  1. 

Evidently  the  rth  factor  in  the  product,  n  {n  —  1)  (?i  —  2)  •  •  • 
is  ?i  —  (r  —  1),  or  n  —  r  +  1.  Hence  the  formula  (1)  may  be 
written 

p»  =  n  (n  -  1)  (^  -x2)  ...(«-  r  +  1).  (2) 

When  r  =  n,  the  factor  n  —  r  -\- 1  is  n  —  n  +  1,  or  1,  and 
we  have  P;;  =  n  (n  —  1)  •  •  •  2  •  1,  or  1  ■  2  ■  •  •  (/i  —  1)  n.  The  con- 
tinued product  1-2  ■  •  ■  n  is  called  factorial  n  and  is  denoted 
by  the  symbol  n !  or  [n^.  Hence  the  efi^ire  number  of  orders 
in  which  we  can  arrange  n  letters  in  a  row,  using  all  of  them 
in  each  arrangement,  is  given  by  the  formula 

Pl  =  nl  (3) 

For  a  reason  which  will  appear  later,  §  775,  the  meaningless 
symbol  0!  is  assigned  the  value  1. 

Example  1.  How  many  different  signals  can  be  made  with  four  flags 
of  different  colors  displayed  singly,  or  one  or  more  together,  one  above 
another  ? 

There  will  be  one  signal  for  each  arrangement  of  the  flags  taken  1,  2,  3, 
or  4  at  a  time.     Hence  the  number  is  Pj  +  P|  -(-  P3  +  P^,  or  64. 

Example  2.  Of  the  permutations  of  the  letters  of  the  word  fancies 
taken  all  at  a  time, 

(1)  How  many  begin  and  end  with  a  consonant  ? 

The  first  place  may  be  filled  in  4  ways,  then  the  last  place  in  3  ways, 
then  the  intermediate  places  in  5 !  ways.  Hence  the  required  number  is 
4  •  3  •  5  !,  or  1440. 

(2)  How  many  have  the  vowels  in  the  even  places  ? 

The  vowels  may  be  arranged  in  the  even  places  in  3  !  ways,  the  con- 
sonants in  the  odd  places  in  4  !  ways,  and  each  arrangement  of  vowels  may 
be  associated  with  every  arrangement  of  consonants.  Hence  the  required 
number  is  3  !  •  4  !,  or  144. 


396  A   COLLEGE   ALGEBRA 

(3)  How  many  do  not  have  c  as  their  middle  letter? 

Evidently  c  is  the  middle  letter  in  0 !  of  the  permutations,  for  the 
remaining  letters  may  be  arranged  in  all  possible  orders.  Therefore  the 
number  of  the  pernmtatious  in  which  c  is  not  the  middle  letter  is  7  !  —  6  !, 
or  4320. 

Example  3.     Show  that  P^  =  4  ■  Pl,  and  that  P^^  -2-  P^. 

Example  4.     If  P\"  =  127  Pl",  find  n. 

Example  5.  How  many  passenger  tickets  will  a  railway  company 
need  for  use  on  a  division  on  which  there  are  twenty  stations? 

Example  6.  In  how  many  of  the  permutations  of  the  letters  a,  e,  i,  o,  u,  y,. 
taken  all  at  a  time,  do  the  letters  a,  e,  i  stand  together  ?       .. 

Example  7.  With  the  letters  of  the  word  numerical  how  many  arrange- 
ments of  five  letters  each  can  be  formed  in  which  the  odd  places  are 
occupied  by  consonants? 

Example  8.     Show  that  with  the  digits  0,  1,  2,  •  •  • ,  9  it  is  possible  to 
(  form  P'^"  —  Pg  numbers,  each  of  which  has  four  different  figures. 

Example  9.  How  many  numbers  all  told  can  be  formed  with  the 
digits  3,  4,  5,  7,  8,  all  the  figures  in  each  number  being  different  ? 

Example  10.  In  how  many  ways  can  seven  boys  be  arranged  in  a  row 
if  one  particular  boy  is  not  permitted  to  stand  at  either  end  of  the  row  ? 

764  Circular  permutations.  The  number  of  different  orders  in 
which  n  different  letters  can  be  arranged  about  the  circum- 
ference of  a  circle  or  any  other  closed  curve  is  (n  —  1)!. 

For  the  relative  order  of  the  n  letters  will  not  be  changed 
if  we  shift  all  the  letters  the  same  number  of  places  along  the 
curve.  Hence  Ave  shall  take  account  of  all  the  distinct  orders 
of  the  n  letters  if  we  suppose  one  of  the  letters  fixed  in  posi- 
tion and  the  remaining  n  —  1  then  arranged  in  all  possible 
orders.  But  these  n  —  1  letters  can  be  arranged  in  {ii  —  1)! 
orders,  §  763,  (3). 

Thus,  eight  persons  can  be  seated  at  a  round  table  in  7  !,  or  5040, 
orders. 

Exam])le  1.  Show  that  the  number  of  circular  r-permutations  of 
n  different  letters  is  P'^/r. 


PERMUTATIONS   AND   COMBINATIONS  397 

Example  2.  Taking  account  of  the  fact  that  a  circular  ring  will  come 
into  coincidence  with  itself  if  revolved  about  a  diameter  through  an  angle 
of  180°,  show  that  (n  —  1)  !/2  different  necklaces  can  be  formed  by  stringing 
.  together  n  beads  of  different  colors. 

Example  3.  In  how  many  ways  can  a  party  of  four  ladies  and  four 
gentlemen  be  arranged  at  a  round  table  so  that  the  ladies  and  gentlemen 
may  occupy  alternate  seats? 

Permutations  of  different  letters  when  repetitions  are  allowed.     765 

With  n  different  letters  we  can  form  n''  arrangements  or  per- 
mutations of  r  letters  each,  if  allowed  to  repeat  a  letter  within 
a  permutation. 

For  in  forming  a  permutation  of  this  kind  we  may  choose 
any  one  of  the  n  letters  for  its  first  letter,  and  then  again, 
since  repetitions  are  allowed,  any  one  of  the  n  letters  for  its 
second  letter,  and  so  on.  Hence,  §  TG2,  the  entire  number  of 
ways  in  which  we  can  form  tlie  permutation  is  »  • « •  n  •  •  •  to 
r  factors,  or  if. 

Thus,  with  the  digits  1,  2,  .3,  •  •  -,  9  we  can  form  all  told  93,  or  729, 
numbers  of  three  figures  each.  - — 

Example  1.  How  many  numbers  of  one,  two,  or  three  figures  each 
can  be  formed  with  the  characters  1,  2,  3,  5,  7  ? 

Example  2.  In  how  many  ways  can  three  prizes  be  given  to  seven 
boys  if  each  boy  is  eligible  for  e\  ery  prize  ? 

The  number  of  n-permutations  of  n  letters  which  are  not  all     766 
different.     Let  us  inquire  how  many  distinguishable  permuta- 
tions can  be  formed  with  the  letters  a,  a,  a,  b,  c(l),  three  of 
which  are  alike,  all  the  letters  being  used  in  each  permutation. 

Compare  these  permutations  with  the  corresponding  permu- 
tations of  the  letters  a,  a',  a",  b,  c(2),  all  of  which  are  different. 
If  we  take  any  one  of  the  permutations  of  (1),  as  abaca,  and, 
leaving  b,  c  undisturbed,  we  interchange  the  a's,  we  get  nothing 
new.  But  if  we  treat  the  corresponding  permutation  of  (2), 
namely  aba'ra",  in  a  similar  manner,  we  obtain  3 !  distinct  per- 
mutations, namely  aba'ca",  aba"ca',  a'ba"ca,  a'baca",  a"haca', 
a"ba'ca.     Hence  to  each  permutation  of  (1)  there  correspond 


398  A   COLLEGE   ALGEBRA 

3 !  permutations  of  (2).  The  number  of  the  permutations  of 
(2)  is  5 !,  §  763,  (3).  Therefore  the  number  of  the  permutations 
of  (1)  is  5! -3!. 

By  the  reasoning  here  employed  we  can  prove  in  general 
that  the  number  of  distinguishable  ?i-permutations  of  n  letters 
of  which  p  are  alike,  q  others  alike,  and  so  on,  is  given  by  the 
formula  , 

2)\q\--- 

Example  1.  In  how  many  different  ways  can  the  letters  of  the  word 
independence  be  arranged  ? 

Of  the  12  letters  in  this  word  4  are  e's,  3  are  n's,  2  are  d's. 

Hence  the  required  result  is  12  !/4  !  •  3  !  •  2  !,  or  1,663,200. 

Example  2.  In  how  many  ways  can  the  letters  of  the  word  Antioch 
be  arranged  without  changing  the  relative  order  of  the  vowels  or  that  of 
the  consonants  ? 

From  the  proof  just  given  it  is  evident  that  the  required  number  of 
arrangements  is  the  same  as  it  would  be  if  the  three  vowels  were  the  same 
and  the  four  consonants  were  the  same.     Hence  it  is  7  !/3  !  •  4  !,  or  35. 

Example  3.  How  many  terms  has  each  of  the  following  symmetric 
functions  of  five  variables  x,  y,  z,  u,  v,  namely,  Sx^y^z,  Zx^y'^z^,  Zx^yzu, 
and  Xx^y^z^u^  ? 

We  shall  obtain  all  the  terms  of  Hx^y'^z  once  each  if,  leaving  the  expo- 
nents 3,  2,  1  fixed  in  position,  we  write  under  them  every  3-permutation 
of  the  letters  x,  y,  z,  u,  v.     Hence  the  number  of  the  terms  is  P^,  or  60. 

If  we  apply  the  same  method  to  :^x^y-z'^,  we  obtain  the  term  x^y-z^ 
twice,  once  in  the  form  x^y^z^  and  once  in  the  form  x^z-y'^.  Similarly 
every  term  is  obtained  twice,  once  for  each  of  the  orders  in  which  its 
letters  under  the  equal  exponents  can  be  written.  Hence  the  number  of 
terms  in  'Lx^y-z'^  is  Pi/ 2,  or  30. 

Similarly  -Zx^yzu  has  P]/3 !,  or  20,  terms,  and  Sx^^/V-it^  has  P^/2  !2  !, 
or  30,  terras. 

Example  4.  In  how  many  ways  can  five  pennies,  six  five-cent  pieces, 
and  four  dimes  be  distributed  among  fifteen  children  so  that  each  may 
receive  a  coin  ? 

Example  5.  In  a  certain  district  of  a  town  there  are  ten  streets  run- 
ning north  and  south,  and  five  running  east  and  west.  In  how  many 
ways  can  a  person  walk  from  the  southwest  corner  of  the  district  to  the 
northeast  corner,  always  taking  the  shortest  course  ? 


PERMUTATIONS   AND    COMBINATIONS  399 

The  number  of  r-combinations  of  n  different  letters.     This  niim-     767 
ber,  C",  is  given  by  the  forinvila 

For  evidently  if  wq  were  to  form  all  the  r-combinations  and 
were  then  to  arrange  the  letters  of  each  combination  in  turn 
in  all  possible  orders,  we  should  obtain  all  the  ^--permutations. 

But  since  each  combination  would  thus  yield  r!  permuta- 
tions, §  763,  (3),  all  the  combinations,  C"  in  number,  would 
yield  r !  x  C"  permutations. 

Hence  r !  x  C"  =  P",  and  therefore  C"  =  P"  h-  r !. 

Thus,  the  numbers  of  combinations  of  the  letters  a,  6,  c,  d,  e  one,  two, 
three,  four,  five  at  a  time  are 

^      r      ^       1.2'      ^       1.2-3'      *       1.2.3-4'      '      1.2.3.4-5" 

This  expression  for  C"  is  the  coefficient  of  the  (r  -f  l)th  term 
in  the  expansion  of  (a  -f-  by  by  the  binomial  theorem,  §  565. 
This  was  shown  in  §  560  by  an  argument  which  is  merely 
another  proof  of  the  formula  (1). 

If  in  the  expression  just  obtained  for  C"  we  multiply  both     768 
numerator  and  denominator  by  (>i  —  r) !,  we  obtain  the  more 
symmetrical  formula 

From  this  formula  (2)  it  follows  that  the  number  of  the     769 
r-combinations  of  n  letters  is  the  same  as  the  number  of  the 

{11  —  ?-)-combinations. 

n  !  n ! 

or  C,\_^  =  ^;;;3-;yr^^^;3-^^^3^^  =  ^^737^177  ^  ^"• 

This  also  follows  from  the  fact  that  for  every  set  of  ?'-things 
taken,  a  set  of  /i  —  r  things  is  left. 

Thus,  C\t  -  CV  =  14  •  13/1  •  2  =  91.  Observe  how  much  more  readily 
C}j  is  found  in  this  way  than  by  a  direct  application  of  (1). 


400  A   COLLEGE    ALGEBRA 

Example  1.  There  are  fifteen  points  in  a  plane  and  no  tlu'ee  of  these 
points  lie  in  the  same  straight  line.  Find  the  number  of  triangles  which 
can  be  formed  by  joining  them. 

Evidently  there  are  as  many  triangles  as  there  are  combinations  of  the 
points  taken  three  at  a  time.  Hence  the  number  of  triangles  is  C'3 ,  that 
is,  15 -14 -13/ 1-2.  3,  or  455. 

Example  2.  In  how  many  ways  can  a  committee  of  three  be  selected 
from  ten  persons  H)  so  as  always  to  include  a  particular  person  A  ?  (2)  so 
as  always  to  exclude  A  ? 

(1)  The  other  two  members  of  the  committee  can  be  chosen  from  the 
remaining  nine  persons  in  C^,  that  is,  9  •  8/1  •  2,  or  36,  ways. 

(2)  The  entire  committee  can  be  chosen  from  the  remaining  nine 
persons  in  C^,  that  is,  9  •  8  •  7/1  •  2  •  3,  or  84,  ways. 

Example  3.  With  the  vowels  a,  e,  i,  0  and  the  consonants  6,  c,  d,  /,  g 
how  many  arrangements  of  letters  can  be  made,  each  consisting  of  two 
vowels  and  three  consonants? 

The  vowels  for  the  arrangement  can  be  chosen  in  C\  ways,  the  con- 
sonants in  Cgways;  then  each  selection  of  vowels  can  be  combined  with 
every  selection  of  consonants  and  the  whole  arranged  in  5  !  ways.  Hence 
the  required  result  is  C|  •  C3  •  5  !,  or  7200. 

Example  4.  In  how  many  ways  can  eighteen  books  be  divided  equally 
among  three  persons  A,  B,  C  ? 

A's  books  can  be  selected  in  C^,?  ways,  then  B's  in  C^^  ways,  then 
C's  in  C;i,  or  1,  way.  Hence,  §702,  the  required  result  is  C^^  ■  C'J  •  C% 
or  18!/(6!)3, 

To  find  in  how  many  ways  the  18  books  can  be  distributed  into  three 
sets  of  6  books  each,  we  must  divide  the  result  just  obtained  by  3  !,  which 
gives  18  !/(()  !)3  3  ! ;  for  liere  the  order  in  which  the  three  sets  may  chance 
to  be  arranged  is  immaterial. 

Example  5.  "With  the  letters  of  the  word  mathematical  how  many 
different  selections  and  how  many  different  arrangements  of  four  letters 
each  can  be  made  ? 

As  the  letters  are  not  all  different  wp  cannot  obtain  the  required  results 
by  single  applicatii)ns  of  the  fornuila.s  f(u-  C"  and  P'^. 

The  letters  are  a,  n,  a ;  ?»,  m  ;  t,  t;  h,  e,  i,  c,  I. 

Hence  we  may  classify  and  then  enumerate  the  possible  selections  and 
arrangements  as  follows : 


PERMUTATIONS    AND    COMBINATIONS  401 

1.  Those  having  three  like  letters. 

Combining  the  3  a's  with  each  of  the  seven  other  letters  in  turn,  we 
obtain  7  selections  and  7  •  4  !/3  !,  or  28,  arrangements. 

2.  Those  having  two  pairs  of  like  letters. 

There  are  3  such  selections  and  3  •  4  !  /  2  !  •  2  !,  or  18,  such  arrangements. 

3.  Those  having  two  letters  alike,  the  other  two  different. 

Of  such  selections  there  are  3  •  C^,  or  63 ;  of  arrangements  there  are 
63-4I/2!,  or  756. 

4.  Those  having  four  different  letters. 

Of  selections  there  are  C%  or  70 ;  of  arrangements,  70  •  4  !,  or  1680. 
Hence  the  total  number  of  selections  is  7  +  3  +  63  +  70,  or  143 ;  of 
arrangements,  28  +  18  +  756  +  1680,  or  2482. 

Example  6.     Find  the  values  of  Cj^,  C'!*,  and  Cj;]. 

Example  7.     If  C^  =  C'-',  find  n. 

Example  8.     If  2  C]  =  5  •  q,  find  n. 

Example  9.  How  many  planes  are  determined  by  twelve  points,  no 
four  of  which  lie  in  the  same  plane  ? 

Example  10.  How  many  parties  of  five  men  each  can  be  chosen  from  a 
company  of  twelve  men  ?  In  how  many  of  these  parties  will  a  particular 
man  A  be  included  ?     From  how  many  will  A  be  excluded  ? 

Example  11.     Of   the   parties   described    in    the    preceding   example 
how  many  will  include  two  particular  men  A  and  B  ?     How  many  wilL— -^ 
include  one  but  not  both  of  them  ?     How  many  will  include  neither  of 
them  ? 

Example  12.  From  twenty  Republicans  and  eighteen  Democrats  how 
many  committees  can  be  chosen,  each  consisting  of  four  Republicans  and 
three  Democrats  ? 

Example  13.  "With  five  vowels  and  fourteen  consonants  how  many 
arrangements  of  letters  can  be  formed,  each  consisting  of  three  vowels 
and  four  consonants? 

Example  14.  In  how  many  ways  can  a  pack  of  fifty-two  cards 
be  divided  equally  among  four  players  A,  B,  C,  D  ?  In  how  many 
ways  can  the  cards  be  distributed  into  four  piles  containing  thirteen 
each  ? 

Example  15.  How  many  numbers,  each  of  five  figures,  can  be  formed 
with  the  characters  2,  3,  4,  2,  5,  2,  3,  6,  7  ? 


402  A    COLLEGE    ALGEBRA 

770  Total  number  of  combinations.  If  in  the  formula  for  (a  +  by, 
§  561,  we  set  a  =  b  =^1  and  then  subtract  1  from  both  mem- 
bers,  we  obtain 

CI+CI  +  ---+  CI  =  2"  -  1. 

Hence  the  total  number  of  combinations  of  n  different 
things  taken  one,  two,  •••,  n  at  a  time,  in  other  words,  the 
total  number  of  ways  in  which  one  or  more  things  may  be 
chosen  from  n  things,  is  2"  —  1. 

This  may  also  be  proved  as  follows  :  Each  particular  thing 
can  be  dealt  with  in  one  of  two  ways,  that  is,  be  taken  or  left. 
Hence  the  total  number  of  ways  of  dealing  with  all  n  things  is 
2-2  ■•-.to  n  factors,  or  2",  §  762.  Therefore,  rejecting  the  case 
in  which  all  the  things  are  left,  we  have  as  before,  2"  —  1. 

Example  1.  How  many  different  sums  of  money  can  be  paid  with  one 
dime,  one  quarter,  one  half  dollar,  and  one  dollar  ? 

Example  2.  By  the  reasoning  just  illustrated,  show  that  the  total  num- 
ber of  ways  in  which  one  or  more  things  can  be  chosen  from  p  +  q  +  ■  •  • 
things  of  which  p  are  alike,  q  others  alike  but  different  from  the  p  things, 
and  so  on,  is  {p  +  1)  (q  +  !)•  •  •  —  1. 

i.^     Example  3.     How  many  different  sums  of  money  can  be  paid  with 
two  dimes,  five  quarters,  and  four  half  dollars? 

771  Greatest  value  of  C°.  In  the  expression  for  C",  namely 
n(?i  —  1)  •  •  •  (?i  —  r  +  1)  /rl,  the  r  factors  of  the  numerator 
decrease  while  those  of  the  denominator  increase.  Hence  for 
a  given  value  of  w  the  value  of  C"  will  be  greatest  when  the 
next  greater  value  of  r  will  make 

(n-r-\-l)/r<l. 

From  this  it  readily  follows  that  if  n  be  even,  C"  is  greatest  when ' 
r  =  n/2  ;  and  if  n  be  odd,  C"  is  greatest  when  r  =  (n  —  l)/2 
or  r=(n-{-l)/2,  the  value  of  C"  being  the  same  for  these 
two  values  of  r,  §  769. 

Example.     What  is  the  greatest  value  of  C;  ?  of  C'/? 


PERMUTATIONS   AND   COMBINATIONS  403 

Combinations  when  repetitions  are  allowed.     Let  us  inquire  in     772 
how  many  ways  we  can  select  three  of  the  four  digits  1,  2,  3,  4 
when  repetitions  are  allowed. 

As  examples  of  such  selections  we  may  take  111,  112,  124, 
illustrating  respectively  the  cases  in  which  all  three,  two,  none 
of  the  digits  are  the  same. 

^  If  to  the  digits  in  111,  in  112,  and  in  124  we  add  0,  1,  2 
respectively,  we  obtain  123, 124,  and  136,  which  are  three  of  the 
3-combinations  tvithout  repetitions  of  the  digits  1,  2,  3,  4,  5,  6. 
And  a  little  reflection  will  show  that  if  we  make  out  a  com- 
plete list  of  the  selections  like  111,  112,  124,  arranging  the 
digits  in  each  so  that  no  digit  is  followed  by  one  of  less  value, 
and  then  to  the  digits  in  each  selection  add  0,  1,  2,  we  shall 
obtain  once  and  but  once  every  one  of  the  3-combinations 
without  repetitions  of  the  4  +  (3  —  1)  or  6  digits  1,  2,  3,  4,  5,  6. 
The  number  of  the  latter  combinations  is  0%  Hence  Cf  is  the 
number  which  we  are  seeking. 

The  same  reasoning  may  be  applied  to  the  general  case  of 
r-combinations,  with  repetitions,  of  the  n  numbers  1,  2,  •••,  7i. 
And  since  the  numbers  may  correspond  to  n  different  things  of 
any  kind,  we  have  the  theorem  : 

The  number  of  the  x-combinations  with  repetitions  ofn  different 
things  is  the  same  as  the  number  of  the  v-combinations  without 
repetitions  ofii  +  v  —  l  different  tilings,  namely,  C"+r~\  or 
n(n  +  l)---(n  +  r  -  l)/r!. 

Example  1.     How  many  different  throws  can  be  made  with  four  dice  ? 

As  any  one  of  the  faces  marked  1,  2,  3,  4,  5,  6  may  turn  uppermost 
in  the  case  of  one,  two,  three,  or  four  of  the  dice,  the  number  of  possible 
throws  is  the  number  of  4-combinations  with  repetitions  of  1,  2,  3,  4,  5,  6, 
namely,  C^^  or  I'^^O. 

Example  2.  How  many  terms  has  a  complete  homogeneous  poly- 
nomial of  the  rth  degree  in  three  variables  a;,  ?/,  z? 

Evidently  it  has  as  many  terms  as  there  are  products  of  the  rth  degree 
whose  facfors  are  x's,  2/"'s,  or  z's.     Hence  the  number  is 

C^  +  ;-^  =  C'+2=  C'-  +  2^(r  +  l)(r  +  2)/2. 


404  A   COLLEGE   ALGEBRA 

773  Formulas  connecting  numbers  of  combinations.  The  correspond- 
ing algebraic  identities.  The  following  relations  are  of  special 
interest  and  importance. 

C«  ^  C'V'  +  C";i}.  (1) 

For  we  may  distribute  the  r-combinations  of  n  letters  into 
two  classes,  — those  which  contain  some  particular  letter,  as  a, 
and  those  which  do  not  contain  this  letter.  We  shall  obtain  all 
the  combinations  of  the  first  class,  once  each,  if  we  form  every 
(r  —  l)-combination  of  the  remaining  n  —  1  letters  and  then  add 
a  to  each  of  these  combinations  ;  hence  their  number  is  C'"z}. 
The  combinations  of  the  second  class  are  the  ^--combinations  of 
the  remaining  n  —  1  letters ;  hence  their  number  is  C"~}. 

(jm+n  =  c-;!  -f  r,™!  •  CI  -f  cv^-o  •  Q  H ^  C'l-  c,.i^  +  c,..  (2) 

For  take  any  group  of  m  -f  71  letters  and  separate  it  into 
two  groups,  one  of  m  letters,  the  other  of  n  letters.  We  shall 
take  account  of  all  the  r-combinations  of  the  m  -f  n  letters,  once 
each,  if  we  classify  them  as  follows.     They  consist  of 

(a)  The  r-combinations  of  the  letters  of  the  m-group.  The 
number  of  these  combinations  is  C"'. 

(h)  The  combinations  which  contain  r  —  1  letters  of  the 
m-group  and  one  letter  of  the  «-group.  As  w^e  can  choose 
the  r  —  1  letters  in  C^^i  ways  and  the  one  letter  in  C"  ways, 
the  number  of  combinations  of  this  kind  is  CV!!i  •  C". 

(c)  The  combinations  which  contain  r  —  2  letters  of  the 
m-group  and  two  letters  of  the  ?i-group.  As  we  can  choose 
the  r  —  2  letters  in  C^™2  ways  and  the  two  letters  in  C"  ways, 
the  number  of  combinations  of  this  kind  is  r,.™2"  Q-  -^^^^  so 
on,  until  last  of  all  we  reach  the  ?--combinations  of  the  letters 
of  the  w-group,  of  which  there  are  C;!. 

Thus,  C^  =  84  and  Cl  +  ClC\  +  C\C\  -f-  C^  =  10  +  40  -F  30  4-  4  =  84. 

774  If  in  (1)  and  (2)  we  replace  the  several  symbols  C  by  their 
expressions  in  terms  of  m,  n,  r,  §  7G7,  (1),  we  obtain*  formulas 
connecting  m,  n,  r.     The  proofs  just  given  only  show  that 


PERMUTATIONS   AND    COMBINATIONS  405 

these  formulas  hold  good  when  m,  n,  r  denote  positive  integers. 
But  in  fact,  so  far  as  m  and  n  are  concerned,  they  are  true 
algebraic  identities,  holding  good  for  all  values  of  these  letters. 
This  may  be  shown  by  arlgebraic  reduction. 

Thus,  in  the  case  of  (1),  we  have 
fju-i   ,   gn-x  ^  (n  -  1)  (n  -  2)  •  •  •  (n  -  r)       (n  -  1)  (n  -  2)  •  •  •  (n  -  r  +  1) 
'■"^  1.2-..r  1  •  2 .  .  .  (r  -  1)  ' 

^  (?i-l)(n-  2)---(n-    r  +  l)    r^      ^f^i^zl] 
1  •  2  •  •  •  (r  -  1)  '  L  r    \ 

-  »(»  -  "^)  •••("•-  ^  +  1)  _   pn 

\-2---r  ~     '' 

But  it  is  not  necessary  to  make  such  a  reduction  to  prove 
that  these  formulas  are  true  identities.  Thus,  when  expressed 
in  terms  of  m,  n,  r,  each  member  of  (2)  denotes  an  integral 
function  of  m  and  n  whose  degree  with  respect  to  each  of  these 
letters  is  r.  These  two  functions  must  be  identically  equal, 
since  otherwise,  were  we  to  assign  some  particular  integral 
value  to  m,  thus  making  them  functions  of  n  alone,  they  could 
not  be  equal  for  more  than  r  values  of  n,  §  421,  whereas  it  has 
already  been  shown  that  in  reality  they  would  be  equal  for  all 
integral  values  of  n. 

EXERCISE   LXV 

1.  If  there  are  three  roads  leading  from  P  to  Q,  two  from  Q  to  It,  and 
four  from  R  to  S,  by  how  many  routes  can  a  person  travel  from  P  to  S? 

2.  In  how  many  ways  can  a  company  of  five  persons  be  arranged  in 
six  numbered  seats  ? 

3.  If  eight  runners  enter  a  half-mile  race,  in  how  many  ways  can  the 
first,  second,  and  third  places  be  won  ? 

4.  In  how  many  ways  can  a  four-oar  crew  be  chosen  from  ten  oars- 
men and  in  how  many  ways  can  all  these  crews  be  arranged  in  the  boat  ? 

5.  From  a  company  of  one  hundred  soldiers  how  many  pickets  of 
three  men  can  be  chosen  ? 

6.  Five  baseball  nines  wish  to  arrange  a  schedule  of  games  in  which 
each  nine  shall  meet  every  other  nine  three  times.  How  many  games 
must  be  scheduled? 


406  A   COLLEGE   ALGEBRA 

7.  In  how  many  ways  can  the  digits  1,  2,  1,  3,  2,  1,  5  be  arranged, 
all  the  digits  occurring  in  each  arrangement  ? 

8.  Of  the  permutations  of  the  letters  in  the  word  factoring,  taken  all 
at  a  time,  (1)  how  many  begin  with  a  vowel  and  end  with  a  consonant? 
(2)  how  many  do  not  begin  with/?  (3)  how  many  have  vowels  in  the 
first  three  places  ? 

9.  In  how  many  of  the  permutations  just  described  do  the  vowels 
retain  the  order  a,  o,  i?  In  how  many  do  the  consonants  retain  the  order 
/,  c,  t,  r,  ?i,  3  ?  In  how  many  do  both  the  vowels  and  the  consonants 
retain  these  orders? 

10.  With  the  letters  of  the  word  resident  how  many  permutations  of 
five  letters  each  can  be  formed  in  which  the  first,  third,  and  fifth  letters 
are  vowels  ? 

11.  In  how  many  ways  can  a  baseball  nine  be  selected  from  fifteen 
candidates  of  whom  six  are  qualified  to  play  in  the  outfield  only  and  nine 
in  the  infield  only  ? 

12.  In  how  Hiany  ways  can  two  numbers  whose  sum  is  even  be  chosen 
from  the  numbers  1,  2,  3,  8,  9,  10  ? 

13.  How  many  numbers  of  one,  two,  or  three  figures  can  be  formed 
with  the  digits  1,  2,  3,  4,  5,  6,  7  (1)  when  the  digits  may  be  repeated? 
(2)  when  they  may  not  be  repeated  ? 

14.  How  many  odd  numbers,  each  having  five  different  figures,  can  be 
formed  with  the  digits  1,  2,  3,  4,  5,  0  ? 

15.  How  many  odd  numbers  without  repeated  digits  are  there  between 
3000  and  8000  ?     How  many  of  these  are  divisible  by  5  ? 

16.  In  how  many  ways  can  a  person  invite  one  or  more  of  five  friends 
to  dinner  ? 

17.  In  how  many  ways  can  fifteen  apples  be  distributed  among  three 
boys  so  that  one  boy  shall  receive  six,  another  five,  and  another  four? 

18.  In  how  many  ways  can  six  positive  and  five  negative  signs  be 
written  in  a  row  ? 

19.  How  many  numbers  f)f  four  figures  each  can  be  formed  with  the 
characters  1,  2,  3,  2,  3,  4,  2,  4,  5,  3,  6,  7  ? 

20.  From  fifteen  French  and  twelve  German  books  eight  French  and 
seven  German  books  are  to  be  selected  and  arranged  on  a  shelf.  In  how 
many  ways  can  this  be  done  ? 


PERMUTATIONS    AND   COMBINATIONS  407 

21.  From  a  complete  suit  of  thirteen  cards  five  are  to  be  selected 
which  shall  include  the  king  or  queen,  or  both.  In  how  many  ways  can 
this  be  done  ? 

22.  In  how  many  ways  can  four  men  be  chosen  from  five  Americans 
and  six  Englishmen  so  as  to  include  (1)  only  one  Englishman  ?  (2)  at 
least  one  Englishman? 

23.  How  many  parallelograms  are  formed  when  a  set  of  ten  parallel 
lines  is  met  by  another  set  of  twelve  parallel  lines  ? 

24.  Given  n  points  in  a  plane  no  three  of  which  lie  in  the  same  straight 
line,  except  m  which  all  lie  in  the  same  straight  line.  Show  that  the 
number  of  lines  obtained  by  joining  these  points  is  Co  —  €"2  +  1. 

25.  Find  the  number  of  bracelets  that  can  be  formed  by  stringing 
together  five  like  pearls,  six  like  rubies,  and  five  like  diamonds. 

26.  In  how  many  ways  can  ten  persons  be  arranged  at  two  round 
tables,  five  at  each  table  ? 

27.  In  how  many  ways  can  six  ladies  and  five  gentlemen  arrange  a 
game  of  lawn  tennis,  each  side  to  consist  of  one  lady  and  one  gentleman  ? 

28.  In  how  many  ways  can  fifteen  persons  vote  to  fill  a  certain  office 
for  which  there  are  five  candidates  ?  In  how  many  of  these  ways  will  the 
vote  be  equally  divided  among  the  five  candidates  ? 

29.  A  boat  crew  consists  of  eight  men  two  of  whom  are  qualified  to 
row  on  the  stroke  side  only  and  one  on  the  bow  side  only.  In  how  many 
ways  can  the  crew  be  arranged  ? 

30.  How  many  baseball  nines  can  be  chosen  from  eighteen  players  of 
whom  ten  are  qualified  to  play  in  the  infield  only,  five  in  the  outfield 
only,  and  three  in  any  position  ? 

31.  Show  that  the  number  of  permutations  of  six  different  letters 
taken  all  at  a  time,  when  two  of  the  letters  are  excluded  each  from  a 
particular  position,  is  6  !  —  2  •  5  !  +  4  !. 

32.  How  many  combinations  four  at  a  time  can  be  formed  with  the 
letters  p,  q,  r,  s,  t,  v,  when  repetitions  are  allowed  ? 

33.  How  many  different  throws  can  be  made  with  five  dice  ? 

34.  How  many  terms  has  each  of  the  symmetric  functions  'E,x*yh^u., 
^x^y^z^u,  'Zxhj^z'^vP'V,  the  number  of  the  variables  being  ten  ? 

35.  Show  that  the  number  of  terms  in  a  complete  homogeneous  func- 
tion  of  the  nth  degree  In  four  variables  is  (n  +  1)  (n  +  2)  (n  +  3)  /S/ 


408  A   COLLEGE    ALGEBRA 

XXVI.     THE    MULTINOMIAL    THEOREM 

775         Multinomial  theorem.     Let  a  +  b  +  ••■  +  k  denote  any  poly 
nomial,  and  n  a  positive  integer.     Then 

(^a  +  b  +  ---  +  ky  =  2  --r— :  a'^b^  ■  ■  ■  k% 

^  '  al  /3l  ■  ■  ■  kI 

where  the  sum  on  the  right  contains  one  term  for  each 'set  of 
values  of  a,  f3,  ■■  ■,  k  that  can  be  selected  from  0,  1,  2,  ■■■.  n, 
such  that  a  +  /S  +  •  •  •  +  K  =  ?i,  it  being  understood  that  when 
a  =  0,  a!  's  to  be  replaced  by  1,  and  the  like  for  /3,  •  •  •,  k. 

For  (a  +  ^  +  •  •  •  +  ky  denotes  the  continued  product 

{a  +  b  y \-k){a  +  b  ^ \- k)  ■  ■  ■  to  n  factors, 

and  each  of  the  partial  products  obtained  by  actually  carrying 
out  this  multiplication,  without  collecting  like  terms,  has  the 
form :  a  letter  from  the  first  parenthesis,  times  a  letter  from 
the  second,  times  a  letter  from  the  third,  and  so  on. 

But  since  the  letter  selected  from  each  parenthesis  may  be 
any  one  of  the  letters  a,  b,  ■  •■,  k,  a,  list  of  the  products  as  thus 
written  would  also  be  a  complete  list  of  the  ?i-perm\itations  of 
the  letters  a,  b,  ■■•,  k  when  repetitions  are  allowed.  And  if 
«,  /3,  •  •  •,  /c  denote  any  particular  set  of  numbers  0, 1,  •  •  •,  w  whose 
sum  is  n,  there  will  be  in  the  list  as  many  products  in  which 
a  of  the  factors  are  a's,  ji  of  them  J's,  •  •  • ,  k  of  them  /c's,  as 
there  are  n-per mutations  of  n  letters  of  which  a  are  alike,  /3 
others  are  alike,  and  so  on,  namely,  w! /a!  •  /3!  •  •  •  /<!,  §7GG.  And 
since  each  of  these  products  is  equal  to  a"i^  •  ■  •  k",  their  sum  is 

a !  /? !  •  •  •  K ! 

The  binomial  theorem  is  a  particular  case  of  this  theorem. 

Tims,  the  expansion  of  (a  +  b  +  c  +  d  +  e)^  consists  of  terms  of  the 
five  types  abed,  a^bc,  a'^b^,  a%  a*  with  the  coefficients  4!/l!l!l!lI,  or 
24;  4!/2!l  !1  !,  or  12;  4!/2!2!,  or  6 ;  4!/3!l!,  or  4;  4!/4!,  or  1, 
respectively.     Hence,  uniting  terms  of  the  same  type,  V7e  have 

(a  +  6  +  c  +  d  +  e)*  =  Sa*  +  4  Sa^ft  +  6 1,aW-  +  12  I,a^bc  +  24  Sa6cd. 


PROBABILITY  409 


Example.     Find  the  coefficient  of  x^  in  the  expansion  of  (2  +  3x  +  4x2)8. 

8  ' 

The  eeneral  form  of  a  term  of  this  expansion  is 2'^3^4v  x3  +  2  y, 

^  cT  !  /3  !  7  ! 

where  a  +  ^  +  7  =  8  (1),  and  the  terms  required  are  those  for  which 
)3  +  27  =  5  (2).  A  complete  list  of  tlie  solutions  of  (1),  (2)  in  positive 
integers  orO's  is  a,  ^3,  7  =  3,  5,  0 ;  4,  3,  1 ;  5,  1,  2.  Hence  the  required 
coefficient  is 

_?_L  23  .  35  +  -^-—  2*  .  33  .  4  +  -^  25  .  3  •  42,  or  850,752. 
315!  4  131.  5!2! 


EXERCISE    LXVI 

1.  Give  the  expansion   o£  (a  +  6  +  c  +  d)^,  collecting  terms  of  the 
same  type. 

2.  Also  the  expansion  of  (a  +  6  +  c  +  d)^. 

3.  Find  the  coefficients  of  a^h^d-d,  a^¥c*,  and  a^hH'^  in  the  expansion 
of  (a  +  6  +  c  +  d)i2. 

4.  Find  the  coefficient  of  ab'^c^d^  in  the  expansion  of  (a  —  6  +  c  —  dY'^. 

5.  Find  the  coefficient  of  a'^Wc  in  the  expansion  of  (a  +  3  6  +  2  cf. 

6.  Find  the  coefficient  of  x*'  in  the  expansion  of  (1  +  x  +  x^  +  x^)^^. 

7.  Find  the  coefficient  of  x'  in  the  expansion  of  (1  —  x  +  3x2)9. 


XXVII.     PROBABILITY 

SIMPLE   EVENTS 

Probability.  Consider  any  future  event  which,  if  given  a 
trial,  that  is,  an  opportunity  to  happen,  must  happen  or  fail 
to  happen  in  one  of  a  limited  number  of  ways  all  equally 
likely,  that  is,  ways  so  related  tliat  there  is  no  reason  for 
expecting  any  one  of  them  rather  than  any  other.  The  turning 
of  the  ace  uppermost  when  a  die  is  thrown  is  such  an  event. 
For  one  of  the  six  faces  of  the  die  must  turn  uppermost,  and 
there  is  no  reason  for  expecting  any  one  face  to  turn  rather 
than  any  other. 


410  A    COLLEGE   ALGEBRA 

Calling  all  the  equally  likely  ways  in  which  such  an  event 
can  happen  or  fail  the  possible  cases  with  respect  to  the  event, 
the  ways  in  which  it  can  happen  the  favorable  cases,  and  the 
ways  in  which  it  can  fail  the  unfavorable  cases,  we  say  : 

The  probability  or  chance  of  the  event  is  the  ratio  of  the 
number  of  favorable  cases  to  the  entire  number  of  possible  cases, 
favorable  and  unfavorable. 

Hence  if  m  denote  the  number  of  possible  cases,  a  the 
number  of  favorable  cases,  and  p  the  probability,  we  have  by 
definition 

p  —  aim. 

Thus,  the  probability  that  the  ace  will  turn  up  when  a  die  is  thrown  is 
1/6  ;  for  here  m  =  6  and  a  =  1. 

Again,  the  chance  of  drawing  a  white  ball  from  a  bag  known  to  contain 
five  balls,  three  white  and  two  black,  is  3/5. 

Corollary  1.  If  an  event  is  certain  to  happen,  its  probability 
is  1 ;  if  it  is  certain  to  fail,  its  probability  is  0 ;  in  every  other 
case  its  probability  is  a  positive  proper  fraction. 

For  if  the  event  is  certain  to  happen,  there  are  no  ways  in 
which  it  can  fail;  hence  a  =  m  and  a  Jm  —  1.  If  the  event 
is  certain  to  fail,  there  are  no  ways  in  which  it  can  happen ; 
hence  a  =  0  and  a  Jm  —  0.  In  every  other  case  a  is  greater 
than  0  and  less  than  ni,  so  that  a  J ni  is  a  positive  proper 
fraction. 

Corollary  2.  If  the  probability  that  an  event  will  happen  is 
p,  the  probability  that  it  will  not  happ)en  is  1  —  p. 

For  if  a  of  the  ni  possible  cases  favor  the  occurrence  of  the 
event,  the  remaining  m  —  a  cases  favor  its  non-occurrence. 
Hence  the  probability  that  the  event  will  not  happen  is 
(m  —  o)  [m  =  1  —  a/m  =  1  —  p. 

Odds.  If  the  number  of  favorable  cases  with  respect  to  a 
certain  event  is  a  and  the  number  of  unfavorable  cases  is  b, 
we  say,  when  a  >  b,  that  the  odds  are  a  to  i  in  favor  of  the 


PROBABILITY  411 

event ;  when  b  >  a,  that  the  odds  are  b  to  a  against  the  event; 
when  a  =  b,  that  the  odds  are  even  on  tlie  event.  In  the  first 
case  the  probability  of  the  event,  namely,  a  /  (a  +  b),  is  greater 
than  1/2;  in  the  second  it  is  less  than  1/2  ;  in  the  third  it 
is  equal  to  1/2. 

Thus,  if  a  ball  is  to  be  drawn  from  a  bag  containing  five  balls,  three 
white  and  two  black,  the  odds  are  3  to  2  in  favor  of  its  being  white,  and 
3  to  2  against  its  being  black. 

Expectation,     li  jj  denote  the  chance  that  a  person  will  win     780 
a  certain  sum  of  money  M,  the  product  Mp  is  called  the  value 
of  his  expectation  so  far  as  this  sum  M  is  concerned. 

Thus,  the  value  of  the  expectation  of  a  gambler  who  is  to  win  $12  if  he 
throws  an  ace  with  a  single  die  is  $12  x  1/6,  or  $2. 

Examples  of  probability.     In  applying  the  definition  of  proba-     781 
bility,  §  776,  care  must  be  taken  to  reduce  the  possible  cases 
to  such  as  are  equally  likely.     The  following  examples  will 
illustrate  the  need  of  this  precaution. 

Example  1.  If  two  coins  be  tossed  simultaneously,  what  is  the  chance 
that  the  result  will  be  two  heads  ?   two  tails  ?   one  head  and  one  tail  ? 

We  might  reason  thus :  There  are  three  possible  cases,  one  favoring 
the  first  result,  one  the  second,  one  the  third  ;  hence  tlie  chance  of  each 
result  is  1/3. 

But  our  conclusion  would  be  false,  since  the  number  of  equally  likely 
possible  cases  is  not  three  but  four.  For  if  we  name  the  coins  A  and  B 
respectively,  the  equally  likely  cases  are  :  A  head,  B  head  ;  A  tail,  B  tail ; 
A  head,  B  tail;  A  tail,  B  head.  And  since  one  of  these  cases  favors 
the  result  two  heads,  one  the  result  two  tails,  and  two  the  result  one 
head  and  one  tail,'  the  chances  of  these  results  are  1/4,  1/4,  and  2/4 
respectively. 

Example  2.  What  is  the  chance  of  throwing  a  total  of  eight  with 
two  dice  ? 

Here  the  number  of  equally  likely  possible  cases  is  6  •  G,  or  36,  for  any 
face  of  one  die  may  turn  up  with  any  face  of  the  other  die. 

We  have  a  total  of  eight  if  the  faces  which  turn  up  read  2,  6  or  3,  5  or 
4,  4.     But  there  are  two  ways  in  which  2,  6  may  turn  up,  namely  2  on  the 


412  A    COLLEGE   ALGEBRA 

die  A  and  6  on  the  die  B,  or  vice  versa.  Similarly  there  are  two  ways  in 
which  3,  5  may  turn  up.  On  the  contrary,  there  is  but  one  way  in  which 
4,  4  can  turn  up.  Hence  there  are  five  favorable  cases.  Therefore  the 
chance  in  question  is  5/36. 

Example  3.  What  is  the  chance  of  throwing  a  total  of  eight  with  three 
dice  if  at  least  one  die  turns  ace  up  ? 

The  number  of  equally  likely  possible  cases  is  6  •  6  ■  6,  or  216. 

We  have  a  total  of  eight  if  the  faces  which  turn  up  read  1,  1,  6  or  1,  2,  5 
or  1,  3,  4.  But  1,  1,  6  may  turn  up  in  3  !/2  !,  or  3,  ways,  §  766  ;  for  the 
numbers  1,  1,  G  may  be  distributed  among  the  three  dice  in  any  of  the 
orders  in  which  1,  1,  6  can  be  written.  Similarly  1,  2,  6  and  1,  3,  4  may 
each  turn  up  in  3  !,  or  6,  ways.  Hence  there  are  3  +  6  +  6,  or  15,  favor- 
able cases.     Therefore  the  chance  in  question  is  15/216,  or  5/72. 

Example  4.  An  urn  contains  six  white,  four  red,  and  two  black 
balls. 

(1)  If  four  balls  are  drawn,  what  is  the  chance  that  all  are  white  ? 
There  are  as  many  ways  of  drawing  four  white  balls  as  there  are  4-com- 

binations  of  the  six  white  balls  in  the  urn,  namely  C^.  Similarly,  since  the 
urn  contains  twelve  balls  all  told,  the  total  number  of  possible  drawings 
is  C\^.     Hence  the  chance  in  question  is  Cl/C^^,  or  1/33. 

(2)  If  six  balls  are  drawniwhat  is  the  chance  that  three  of  them  are 
white,  two  red,  and  one  black  ? 

The  three  white  balls  can  be  chosen  in  C^  ways,  the  two  red  balls  in  Ct 
ways,  the  one  black  ball  in  Cf  ways.  Hence  the  number  of  ways  in  which 
the  required  drawing  can  be  made  is  C^-  C^-  Cf.  The  total  number  of 
possible  drawings  is  C',?.  Hence  the  chance  in^uestion  is  Cg  •  C|  ■  C\/  Cg^, 
or  20/77. 

Example  5.     Three  cards  are  drawn  from  a  suit  of  thirteen  cards. 

(1 )  What  is  the  chance  that  neither  king  nor  queen  is  drawn  ? 
Aside  from  the  king  and  queen  there  are  eleven  cards.     Hence  there  are 

Cy  sets  of  three  cards  which  include  neither  king  nor  queen.  Therefore 
the  probability  in  question  is  C\j /  C^^,  or  15/26. 

(2)  What  is  the  chance  that  king  or  queen  is  drawn,  one  or  both? 
This  event  occurs  when  the  event  described  in  (1)  fails  to  occur.    Hence 

the  probability  in  question  is  1  -  15/26,  or  11/26,  §  778. 

(3)  Wliat  is  the  chance  that  both  king  and  queen  arc  drawn? 

We  obtain  every  set  of  three  cards  which  includes  both  king  and  queen 
if  we  combine  each  of  the  remaining  eleven  cards  in  turn  with  king  and 
queen.     Hence  the  chance  in  question  is  ll/C'g,  or  1/26. 


PROBABILITY  413 

On  the  various  meanings  of  probability.  1.  The  fraction  o/m,  782 
which  we  have  called  the  probability  of  an  event,  §  776,  means 
nothing  so  far  as  the  actual  outcome  of  a  single  trial,  or  a 
small  number  of  trials,  of  the  event  is  concerned.  But  it  does 
indicate  the  frequency  with  which  the  event  would  occur  in 
the  long  run,  that  is,  in  the  course  of  an  indefinitely  long 
series  of  trials. 

Thus,  if  one  try  the  experiment  of  throwing  a  die  a  very 
great  number  of  times,  say  a  thousand  times,  he  will  find  that 
as  the  number  of  throws  increases  the  ratio  of  the  number  of 
times  that  ace  turns  up  to  the  total  number  of  throws  approaches 
the  value  1/6  more  and  more  closely. 

2.  There  are  important  classes  of  events  —  the  duration  of 
life  is  one  —  to  which  the  definition  of  §  776  does  not  apply, 
it  being  impossible  to  enumerate  the  ways,  all  equally  likely, 
in  which  the  event  can  happen  or  fail.  But  we  may  be  able  to 
determine  the  frequency  with  which  events  of  such  a  class  have 
occurred  in  the  course  of  a  very  great  number  of  ^>asi^  trials. 
If  so,  we  call  the  fraction  which  indicates  this  frequency  the 
probability  of  an  event  of  the  class.  Like  1/6  in  the  case  of 
the  die,  it  indicates  the  frequency  with  which  events  of  the 
class  may  reasonably  be  expected  to  occur  in  the  course  of  a 
very  great  number  of  future  trials. 

Thus,  if  we  had  learned  from  the  census  reports  that  of 
100,000  persons  aged  sixty  in  1880  about  2/3  were  still  living 
in  1890,  we  should  say  that  the  probability  that  a  person  now 
sixty  will  be  alive  ten  years  hence  is  2/3. 

3.  But  we  also  use  the  fraction  a/ m  to  indicate  the  strength 
of  our  expectation  that  the  event  in  question  will  occur  on 
a  single  trial.  The  greater  the  ratio  of  the  number  of  favor- 
able cases  to  the  number  of  possible  cases,  or  the  greater  the 
frequency  with  which,  to  our  knowledge,  events  of  a  similar 
character  have  occurred  in  the  past,  the  stronger  is  our  expec- 
tation that  this  particular  event  will  occur  on  the  single  trial 
under  consideration. 


414  A   COLLEGE   ALGEBRA 

We  may  speak  of  the  probability,  in  this  sense,  of  any  kind 
of  future  event.  Thus,  before  a  game  between  two  football 
teams,  A  and  B,  we  hear  it  said  that  the  odds  are  3  to  2  in 
favor  of  A's  winning,  or  that  the  probability  that  A  will  win  is 
3/5.  This  means  that  the  general  expectation  of  A's  winning 
is  about  as  strong  as  one's  expectation  of  drawing  a  white  ball 
from  an  urn  known  to  contain  five  balls  three  of  which  are 
white. 

EXERCISE  LXVII 

1.  The  probability  of  a  certain  event  is  3/8.  Are  the  odds  in  favor 
of  the  event  or  against  it,  and  what  are  these  odds  ?  What  is  the  prob- 
ability that  the  event  will  not  occur  ? 

2.  The  odds  are  10  to  9  in  favor  of  A's  winning  a  certain  game. 
What  is  the  chance  of  his  winning  the  game  ?    of  losing  it  ? 

3.  The  odds  are  5  to  3  in  favor  of  A's  winning  a  stake  of  $G0.  What 
is  his  expectation  ? 

4.  The  French  philosopher  D'Alembert  said  :  "There  are  two  possible 
cases  with  respect  to  every  future  event,  one  that  it  will  occur,  the  other 
that  it  will  not  occur.  Hence  the  chance  of  every  event  is  1  /2  and  the 
definition  of  probability  is  meaningless."     How  should  he  be  answered  ? 

5.  An  urn  contains  sixteen  balls  of  which  seven  are  white,  six  black, 
and  three  red. 

(1)  If  a  single  ball  be  drawn,  what  is  the  chance  that  it  is  white? 
black?   red? 

(2)  If  two  balls  be  drawn,  what  is  the  chance  that  both  are  black? 
one  white  and  one  red  ? 

(3)  If  three  balls  be  drawn,  what  is  the  chance  that  all  are  red? 
none  red?   one  white,  one  black,  one  red? 

(4)  If  four  balls  be  drawn,  what  is  the  chance  that  one  is  white  and 
the  rest  not  ?   two  white  and  the  other  two  not  ? 

(5)  If  ten  balls  be  drawn,  what  is  the  chance  that  five  are  white,  three 
black,  and  two  red  ? 

6.  What  is  the  chance  of  throwing  doublets  with  two  dice  ?   with  > 
three  dice  ? 

7.  What  is  the  chance  of  throwing  a  total  of  seven  with  two  dice  ? 
Show  that  this  is  the  most  probable  throw. 


PROBABILITY  415 

8.  What  is  the  chance  of  throwing  at  least  one  ace  in  a  throw  with 
two  dice  ?   of  throwing  one  ace  and  but  one  ? 

9.  One  letter  is  taken  at  random  from  each  of  the  words  factor  and 
banter.     What  is  the  chance  that  the  same  letter  is  taken  from  each  ? 

10.  A  box  contains  nine  tickets  numbered  1,  2,  •  •  • ,  9.  If  two  of  the 
tickets  be  drawn  at  random,  what  is  the  chance  that  the  product  of  the 
numbers  on  them  is  even  ?   odd  ? 

i^'ll.  If  five  tickets  be  drawn  from  this  box,  find  the  chance  (1)  that 
1,  2,  and  3  are  drawn  ;  (2)  that  one  and  but  one  of  1,  2,  and  3  is  drawn  ; 
(3)  that  none  of  these  numbers  is  drawn. 

12.  If  four  cards  be  drawn  from  a  complete  pack  of  fifty-two  cards, 
what  is  the  chance  that  they  are  ace,  king,  queen,  and  knave  ?  ace,  king, 
queen,  and  knave  of  the  same  suit  ? 

I  J:3.  What  is  the  chance  that  a  hand  at  whist  contains  four  trumps  and 
three  cards  of  each  of  the  remaining  suits  ? 

14.  What  is  the  chance  of  a  total  of  five  in  a  single  throw  with  three 
dice  ?   of  a  total  of  less  than  five  ? 

15.  If  eight  persons  be  seated  at  a  round  table,  what  is  the  chance  that 
two  particular  persons  sit  together  ? 


COMPOUND  EVENTS.     MUTUALLY  EXCLUSIVE  EVENTS 

Independent  events.     Two  or  more  events  are  said  to  be  inde-     783 
pendent  when  the  occurrence  or  non-occurrence  of  any  one 
of  them  is  not  affected  by  the  occurrence  or  non-occurrence  of 
the  rest.     In   the  contrary   case  the    events   are   said  to  be 
interdependent. 

Thus,  the  results  of  two  drawings  of  a  ball  from  a  bag  are  independent 
if  the  ball  is  returned  after  the  first  drawing,  but  interdependent  if  the 
ball  is  not  returned. 

Theorem  1.     The  prohabiUty  that  all  of  a  set  of  independent     784 
events  will  occur  is  the  product  of  the  j)robahilities  of  the  single  _-> 
events.  '" 

For  consider  two  such  events  whose  probabilities  are  «i/mi 
and  ag/wg  respectively. 


416  A    COLLEGE   ALGEBRA 

The  number  of  equally  likely  possible  cases  for  and  against 
the  first  event  is  ?»i,  for  and  against  the  second  vi^,  and  since 
the  events  are  independent  any  one  of  the  m^  cases  may  occur 
with  any  one  of  the  w^  cases.  Hence  the  number  of  equally 
likely  possible  cases  for  and  against  the  occurrence  of  both 
events  is  inim<i.  And  by  the  same  reasoning,  a^a^  of  these 
cases  favor  the  occurrence  of  both  events.  Therefore  the  prob- 
ability that  both  events  will  occur  is  ^  "  »  that  is  -^  ■  -^  >  as 
was  to  be  demonstrated. 

The  proof  for  the  case  of  more  than  two  events  is  similar. 

The  demonstration  applies  only  to  events  of  the  kind 
described  in  §  776,  but  for  the  reasons  indicated  in  §  782  we 
may  apply  the  theorem  itself  to  any  kind  of  future  event,  as 
in  Ex.  2  below. 

Thus,  the  chance  of  throwing  ace  twice  in  succession  with  a  single  die 
is  1/6  X  1/6,  or  1/36. 

Again,  the  chance  of  twice  drawing  a  white  ball  from  a  bag  which 
contains  five  white  and  four  black  balls,  the  ball  first  drawn  being  returned 
before  the  second  drawing,  is  5/9  x  5/9,  or  25/81. 

Theorem  2.  If  the  probability  of  a  first  event  is  pi,  and  if 
after  this  event  has  happened  the  probahility  of  a  second  event 
is  p2,  the  probability  thai  both  events  will  occur  in  the  order 
stated  is  piP2.     And  similarly  for  more  than  two  events. 

This  theorem  may  be  proved  in  the  same  manner  as 
Theorem  1.     It  evidently  includes  that  theorem. 

Thus,  after  a  white  ball  has  been  drawn  from  a  bag  containing  five 
white  and  four  black  balls,  and  not  replaced,  the  chance  of  drawing  a 
second  white  ball  is  4/8.  Hence  the  chance  of  twice  drawing  a  white 
ball  when  the  one  first  drawn  is  not  replaced  is  5/9  x  4/8,  or  5/18. 

Example  1.  What  is  the  chance  that  ace  will  turn  up  at  least  once  in 
the  course  of  three  throws  with  a  die  ? 

Ace  will  turn  up  at  least  once  unless  it  fails  to  turn  in  every  throw. 
The  chance  of  failure  in  a  single  throw  being  5/6,  the  chance  of  failure  in 
all  three  throws  is  5/6  x  5/6  x  5/0,  or  125/216.  Hence  the  chance  of 
at  least  one  ace  is  1  -  125/216,  or  91/216. 


PROBABILITY  417 

Example  2.  The  chance  that  A  will  solve  a  certain  problem  is  3/4. 
The  chance  that  B  will  solve  it  is  2/3.  What  is  the  chance  that  the 
problem  will  be  solved  if  both  A  and  B  attempt  it  independently  ? 

The  problem  will  be  solved  unless  both  A  and  B  fail.  The  chance  of 
A's  failure  is  1/4,  of  B's  failure  1/3.  Hence  the  chance  that  both  fail 
is  1/4  X  1/3,  or  1/12.  Therefore  the  chance  that  the  problem  will  be 
solved  is  11/12. 

Example  3.  There  are  two  purses,  one  containing  five  silver  coins  and 
one  gold  coin,  the  other  three  silver  coins.  If  four  coins  be  drawn  from 
the  first  purse  and  put  into  the  second,  and  five  coins  be  then  drawn 
from  the  second  purse  and  put  into  the  first,  what  is  the  chance  that  the 
gold  coin  is  in  the  second  purse  ?   in  the  first  purse  ? 

The  chance  that  the  gold  coin  is  taken  from  the  first  purse  and  put 
into  the  second  is  C\/  C%  or  2/3,  §  781,  Ex.  5.  The  chance  that  it  is  then 
left  in  the  second  purse  is  ^  /  C\^  or  2/7.  Hence  the  chance  that  after 
both  drawings  it  is  in  the  second  purse  is  2/3  x  2/7,  or  4/21.  The 
chance  that  it  is  in  the  first  purse  is  1  —  4/21,  or  17/21. 

Example  4.  If  eight  coins  be  tossed  simultaneously,  what  is  the  chance 
that  at  least  one  of  them  will  turn  head  up  ? 

Example  5.  Four  men  A,  B,  C,  and  D  are  hunting  quail.  If  A 
gets  on  the  average  one  quail  out  of  every  two  that  he  fires  at,  B  two 
out  of  every  three,  C  four  out  of  every  five,  and  D  five  out  of  every  seven, 
what  is  the  chance  that  they  get  a  bird  at  which  all  happen  to  fire 
simultaneously  ? 

Example  6.  An  urn  A  contains  five  white  and  four  red  balls.  A  sec- 
ond urn  B  contains  six  white  and  two  black  balls.  What  is  the  chance  of 
drawing  a  white  ball  from  A  and  then,  this  ball  having  been  put  into  B, 
of  drawing  a  white  ball  from  B  also  ? 

Example  7.  The  chance  that  A  will  be  alive  five  years  hence  is  3/4; 
B,  5/6.  What  is  the  chance  that  five  years  hence  both  A  and  B  will  be 
alive  ?   A  alive,  B  dead  ?   A  dead,  B  alive  ?   both  dead  ? 

Mutually  exclusive  events.     If  two  or  more  events  are  so     786 
related  that  but  one  of  them  can  occur,  they  are  said  to  be 
mutually  exclusive. 

Thus,  the  turning  of  an  ace  and  the  turning  of  a  deuce  on  the  same 
throw  of  a  single  die  are  mutually  exclusive  events. 


i   -1^ 


^  a 


418  A    COLLEGE   ALGEBRA 

787  Theorem  3.  The  probabiliti/  that  some  one  or  other,  of  a  set 
of  mutually  exclusive  events  will  occur  is  the  sum  of  the 
■probabilities  of  the  single  events. 

For  consider  two  mutually  exclusive  events  A  and  B. 

The  possible  cases  with  respect  to  the  two  events  are  of  three 
kinds,  all  mutually  exclusive,  namely,  those  for  which  (1)  A 
happens,  B  fails ;  (2)  A  fails,  B  happens ;  (3)  A  fails,  B  fails. 

Let  the  numbers  of  equally  likely  possible  cases  of  these 
three  kinds  be  I,  m,  and  n  respectively.     Then 

(a)  The  chance  that  either  A  bv  B  happens  is  •; 

For  there  are  I  -\-  m  -\-  n  possible  and  I  +  m  favorable  cases- 

(b)  The  chance  of  the  single  event  A  is ; r- 

1  +  (m  +  n) 

For  since  A  never  happens  except  when  B  fails,  the  I  cases 
in  which  A  happens  and  B  fails  are  all  the  cases  in  which  A 
happens,  and  the  m  +  n  cases  in  which  A  fails  and  B  happens 
or  both  A  and  B  fail  are  all  the  cases  in  which  A  fails. 

(c)  Similarly  the  chance  of  the  single  event  B  is -• 

^  ^  "^  ^  7n-\-(l-\-  n) 

Bt  ^  +  '>n       _  I m 

I  -\-  m  +  n      I  +  {in  +  n)       m  +  (^  +  n) 

Therefore  the  chance  that  either  A  or  B  happens  is  the  sum 
of  the  chances  of  the  single  events  A  and  B. 
The  proof  for  more  than  two  events  is  similar. 

Thus,  if  one  ball  be  drawn  from  a  bag  containing  four  white,  five 
black,  and  seven  red  balls,  since  the  chance  of  its  being  white  is  1/4, 
and  that  of  its  being  black  is  5/16,  the  chance  of  its  being  either  white 
or  black  is  1/4  +  5/16,  or  9/16.  Of  course  this  result  may  be  obtained 
directly  from  the  definition  of  probability,  §  776.  In  fact  that  definition 
may  be  regarded  as  a  special  case  of  Theorem  3. 

Care  must  be  taken  not  to  apply  this  theorem  to  events 
which  are  not  mutually  exclusive. 

Thus,  if  asked,  as  in  §  785,  Ex.  2,  to  find  the  chance  that  a  problem 
will  be  solved  if  both  A  and  B  attempt  it,  A's  chance  of  success  being 


PROBABILITY  419 

3/4  and  B's  2/3,  we  cannot  obtain  the  result  by  merely  adding  3/4  and 
2/3,  since  the  two  events  A  succeeds,  B  succeeds  are  not  mutually  exclu- 
sive. The  mutually  exclusive  cases  in  which  the  problem  will  be  solved 
are :  A  succeeds,  B  fails ;  A  fails,  B  succeeds ;  A  succeeds,  B  succeeds. 
The  chances  of  these  cases  are,  §  784,  3/4x1/3  or  3/12,  1/4x2/3 
or  2/12,  3/4  x  2/3  or  6/12;  and  the  sum  of  these  three  chances,  or 
11/12,  is  the  chance  that  the  problem  will  be  solved. 

Example  1.  An  urn  A  contains  ten  balls  three  of  which  are  white, 
and  a  second  urn  B  contains  twelve  balls  four  of  which  are  white.  If  one 
of  the  urns  be  chosen  at  random  and  a  ball  drawn  from  it,  what  is  the 
chance  that  the  ball  is  white  ? 

We  are  required  to  find  the  chance  of  one  of  the  following  mutually 
exclusive  events  :  (1)  choosing  A  and  then  drawing  a  white  ball  from  it; 
(2)  choosing  B  and  then  drawing  a  white  ball  from  it. 

The  chance  of  choosing  A  is  1  /  2,  and  the  chance  when  A  has  been 
chosen  of  drawing  a  white  ball  is  3/10.  Hence  the  chance  of  (1)  is 
1/2  X  3/10,or3/20.     Similarly  tlie  chance  of  (2)  is  1 /2  x  4/12,orl/6. 

Therefore  the  chance  in  question  is  3/20  +  1/6,  or  19/60. 

Example  2.  What  is  the  value  of  the  expectation  of  a  person  who  is 
to  have  any  two  coins  he  may  draw  at  random  from  a  purse  which 
contains  five  dollar  pieces  and  seven  half-dollar  pieces? 

The  value  of  his  expectation  so  far  as  it  depends  on  drawing  two  dollar 
pieces  is  $2  x  C'^/C'|  =  $2  x  5/33  =  §.30;  on  drawing  two  half-dollar 
pieces,  $1  x  C\/  Clf  =  $1  x  7  /22  =  3-32  ;  on  drawing  one  dollar  piece  and 
one  half-dollar  piece,  .$1.50  x  5  ■  7/  Clf  =  §1.50  x  35/66  =  $.80. 

Hence  the  total  value  of  his  expectation  is  $.30  +  $.32  -|-  .§.80,  or 
$1.42. 

Example  3.  Two  persons  A  and  B  are  to  draw  alternately  one  ball 
at  a  time  from  a  bag  containing  three  white  and  two  black  balls,  the 
balls  drawn  not  being  replaced.  If  A  begins,  what  chance  has  each  of 
being  the  first  to  draw  a  white  ball  ? 

The  chance  that  A  succeeds  in  the  first  drawing  is  3  /  5. 

The  chance  that  A  fails  and  B  then  succeeds  is  2/5  x  3/4,  or  3/10, 
for  when  B  draws,  the  bag  contains  four  balls  three  of  which  are 
white. 

The  chance  that  A  fails,  B  fails,  and  A  then  succeeds  is  2/5x1/4x3/ 3, 
or  1  / 10,  for  when  A  draws,  the  bag  contains  three  balls  all  white. 

Therefore  A's  total  chance  is  3/5 -H 1/10,  or  7/10,  and  B's  is 
3/10. 


420  A    COLLEGE   ALGEBRA 

Example  4.  In  the  drawing  described  in  Ex.  3  what  are  the  respective 
chances  of  A  and  B  if  the  balls  are  replaced  as  they  are  drawn  ? 

On  the  first  round  A's  chance  is  3/5,  B's  2/5  x  3/5,  or  6/25;  and 
their  chances  on  every  later  round,  of  which  there  may  be  any  number, 
will  be  the  same  as  these. 

Hence  their  total  chances  are  in  the  ratio  3/5  :  6/25,  or  5  :  2  ;  that  is, 
A's  total  chance  is  5/7,  B's  2/7. 
"^  Example  5.  In  a  room  there  are  three  tables  and  on  them  nine,  ten, 
and  eleven  books  respectively.  I  wish  any  one  of  six  books,  two  of  which 
are  on  the  first  table,  three  on  the  second,  one  on  the  third.  If  a  friend 
select  a  book  for  me  at  random  from  those  in  the  room,  what  is  the 
chance  that  it  is  one  of  those  I  wish? 
^^f^  Example  6.  An  owner  of  running  horses  enters  for  a  certain  race  two 
hor.ses  whose  chances  of  winning  are  1/2  and  1/3  respectively.  What 
is  the  chance  that  he  will  obtain  the  stakes  ? 

Example  7.  A  and  B  throw  alternately  with  two  dice  for  a  stake 
which  is  to  be  won  by  the  one  who  first  throws  a  doublet.  What  are 
their  respective  chances  of  winning  if  A  throws  first  ? 

788  Repeated  trials  of  a  single  event.  The  following  theorems 
are  concerned  with  the  question  of  the  chance  that  a  certain 
event  will  occur  a  specified  number  of  times  in  the  course  of 
a  series  of  trials,  the  chance  of  its  occurrence  on  a  single  trial 
being  known. 

789  Theorem  4.  If  the  prohahillty  that  an  event  will  occur  on  a 
single  trial  is  p,  the  prohahility  that  it  will  occur  exactly/  v  times 
in  the  course  ofi\  trials  is  C"p''q"~'',  loJiere  q  =  1  —  p. 

For  the  probability  that  it  will  occur  on  all  of  any  partic- 
ular set  of  r  trials  and  fail  on  the  remaining  Ji  —  r  trials  is 
jo''(l  —  p)"~'',  ov  ]j''q"-'\  U  (J  =  1  —J),  §  784. 

But  since  there  are  7i  trials  all  told,  we  may  select  this 
particular  set  of  r  trials  in  C"  ways  which,  of  course,  are 
mutually  exclusive. 

Hence  the  probability  in  (]uestion  is  C'^.p'q"''',  §  787. 

Thus,  the  chance  that  ace  will  turn  up  exactly  twice  in  five  throws  with 
a  single  die,  or  that  out  of  five  dice  thrown  simultaneously  two  and  but 
two  will  turn  ace  up,  is  C^  ■  (!)2  •  {f)3,  or  625/3888. 


PROBABILITY  421 

Observe  that  C",'.p''q"~''  is  the  term  containing/)'"  in  the  expan- 
sion of  (p  +  (/)"  by  the  binomial  theorem ;  lor  C"  =  C,,"^. 

Theorem  5.     The  i^fobabiUty  that  such  an  event  will  occur     790 
at  least  r  times  in  the  course  ofn  trials  is  the  sum  of  the  first 
n  —  r  +  1  terms  in  the  exjjansion  o/"  (p  +  cj)",  namely, 
p„  ^  Cjp^-'q  +  CSp°-2q^  +  •  •  •  +  C„1,p\l"-^ 

For  the  event  will  occur  at  least  r  times  if  it  occurs  exactly 
r  times  or  exactly  any  number  of  times  g)\ieater  than  r,  and 
the  terms  j9",  Cij)"~^q,  •■■,  C„",.^''y"~''repres'^iit  the  probability 
of  the  occurrence  of  the  event  exactly  n,  §i  —  1,  ■  ■  ■,  r  times 
respectively,  §  789.  ! 

Thus,  the  chance  that  ace  will  turn  up  at  least  twice  in  the  course  of 
five  throws  with  a  single  die  is 

a)^  +  5(i)*  I  +  10(DMD^  +  10(i)2(|)3,  or  3^8- 
Example  1.     Two  persons  A  and  B  are  playi;ig  a  game  which  cannot 

be  drawn  and  in  which  A's  skill  is  twice  B's.     What  is  the  chance  that 

A  will  win  as  many  as  three  such  games  in  a  se'.  of  five  ? 

A's  chance  of  winning  a  single  game  is  2/3,  of  losing  1  /3.     Hence  the 

chance  that  A  will  win  as  many  as  three  of  the  five  games  is  the  sum  of  the 

first  three  terms  of  Q  +  })5,  that  is,  (2)5  +  5(2)4  i  +  10(2)3(i)2,  or  64/81. 

Example  2.  Under  the  conditions  of  Ex.  1  what  is  the  chance  that  A 
will  win  three  games  before  B  wins  two  ? 

The  chance  in  question  is  that  of  A's  winning  at  least  three  of  the  first 
four  games  played  ;  and  this  chance  is  (5)*  +  4(|)'^i,  or  J^. 

And,  in  general,  the  chance  of  A's  winning  m  games  before  B  wins  n  is 
the  same  as  the  chance  of  A's  winning  at  least  m  of  the  first  m  +  m  —  1 
games  played. 

Example  3.  Ten  coins  are  tossed  simultaneously.  What  is  the  chance 
that  exactly  six  of  tliem  turn  heads  up  ?   that  at  least  six  turn  heads  up  ? 

Example  4.  If  four  dice  be  thrown  simultaneously,  what  is  the  chance 
that  exactly  three  turn  ace  up?   that  at  least  three  turn  ace  up? 

Example  5.  Under  the  conditions  stated  in  Ex.  1  what  is  the  chance 
that  A  will  win  at  least  four  of  the  five  games  played  ? 

Example  6.  Under  the  same  conditions  what  is  the  chance  that  A  will 
win  four  games  before  B  wins  one  ? 


422  ..   COLLEGE    ALGEBRA 


EXERCISE  LXVra 


1.  A  bag  contailis  three  white,  five  black,  and  seven  red  balls.  On 
the  understanding  that  one  ball  is  dravi-n  at  a  time  and  replaced  as  soon  as 
drav?n,  what  are  the  chances  of  drawing  (1)  first  a  white,  then  a  red,  then 
a  black  ball  ?  (2)  a  white,  red,  and  black  ball  in  any  order  whatsoever  ? 

2.  What  is  the  chance  of  obtaining  a  white  ball  in  the  first  only  of 
three  successive  drawings  from  this  bag,  balls  not  being  replaced  ? 

e 

3.  What  is  the  v  ;lue  of  the  expectation  of  a  person  who  is  allowed  to 
draw  two  coins  at  ra.  dom  from  a  purse  containing  five  fifty -cent  pieces, 
four  dollar  pieces,  ai  (.  three  five-dollar  pieces  ? 

4.  The  chance  tF'it  a  certain  door  is  locked  is  1/2.  The  key  to 
the  door  is  one  of  a  ""launch  of  eight  keys.  If  I  select  three  of  these 
keys  at  random  and  go  to  the  door,  what  is  the  chance  of  my  being  able 
to  open  it  ? 

5.  There  are  three  independent  events  whose  chances  are  1/2,  2/3, 
and  3/4  respectively.  #hat  is  the  chance  that  none  of  the  events  will 
occur?  that  one  and  but  one  of  them  will  occur?  that  two  and  but  two 
will  occur  ?   that  all  thret  will  occur  ? 

6.  Find  the  odds  against  throwing  one  of  the  totals  seven  or  eleven 
in  a  single  throw  with  two  dice. 

7.  What  are  the  odds  against  throwing  a  total  of  ten  with  three  dice  ? 
What  are  the  odds  in  favor  of  throwing  a  total  of  more  than  five  ? 

8.  Three  tickets  are  drawn  from  a  case  containing  eleven  tickets 
numbered  1,  2,  •  •  ■ ,  11.  What  is  the  chance  that  the  sum  of  their  numbers 
Is  twelve  ?     What  is  the  chance  that  this  sum  is  an  odd  number  ? 

9.  Two  gamblers  A  and  B  throw  two  dice  under  an  agreement  that 
if  seven  is  thrown  A  wins,  if  ten  is  thrown  B  wins,  if  any  other  number  is 
thrown  the  stakes  are  to  be  divided  equally.     Compare  their  chances. 

10.  The  same  two  gamblers  play  under  an  agreement  that  A  is  to 
win  if  he  throws  six  before  B  throws  seven,  and  that  B  is  to  win  if  he 
throws  seven  before  A  throws  six.  A  is  to  begin  and  they  are  to  throw 
alternately.     Compare  their  chances. 

11.  Three  gamblers  A,  B,  and  C  put  four  white  and  eight  black  balls 
into  a  bag  and  agree  that  the  one  who  first  draws  a  white  ball  shall  win. 


PROBABILITY  423 

If  they  draw  in  the  order  A,  B,  C,  what  are  their  respective  chances 
when  the  balls  drawn  are  not  replaced  ?   when  they  are  replaced  ? 

12.  What  is  the  worth  of  a  ticket  in  a  lottery  of  one  hundred  tickets 
having  five  prizes  of  $100,  ten  of  §50,  and  twenty  of  $5  ? 

13.  A  bag  A  contains  five  balls  one  of  which  is  white,  and  a  bag  B  six 
balls  none  of  which  is  white.  If  three  balls  be  drawn  from  A  and  put 
into  B  and  three  balls  be  then  drawn  from  B  and  put  into  A,  what  is  the 
chance  that  the  white  ball  is  in  A  ? 

14.  The  bag  A  contains  ??i  balls  a  of  which  are  white,  and  the  bag  B 
contains  n  balls  b  of  which  are  white.  Is  the  chance  of  obtaining  a  white 
ball  by  drawing  a  single  ball  from  one  of  these  bags  chosen  at  random 
the  same  that  it  would  be  if  all  the  balls  were  put  into  one  bag  and  a 
single  ball  then  drawn  ? 

15.  In  a  certain  town  five  deaths  occurred  within  ten  days  including 
January  first.  What  is  the  chance  that  none  of  the  deaths  occurred  on 
January  first? 

16.  If  on  the  average  two  persons  out  of  three  aged  sixty  live  to  be 
seventy,  what  is  the  chance  that  out  of  five  persons  now  sixty  at  least 
three  will  be  alive  ten  years  hence  ? 

17.  A  boy  is  able  to  solve  on  the  average  three  out  of  five  of  the  prob- 
lems set  him.  If  eight  problems  are  given  in  an  examination  and  five 
are  required  for  passing,  what  is  the  chance  of  his  passing  ? 

18.  A  person  is  to  receive  a  dollar  if  he  throws  seven  at  the  first  throw 
with  two  dice,  a  dollar  if  he  throws  seven  at  the  second  throw,  and  so  on 
until  he  throws  seven.     What  is  the  total  value  of  his  expectation  ? 

19.  In  playing  tennis  with  B,  A  wins  on  the  average  three  games  out 
of  four.  What  is  the  chance  that  he  will  win  a  set  from  B  by  the  score 
of  six  to  three  ?  What  is  the  total  chance  of  his  winning  a  set  from  B, 
the  case  of  deuce  sets  being  disregarded  ? 

20.  Under  the  conditions  described  in  the  preceding  example  what 
chance  has  A  of  winning  a  set  in  which  the  score  is  now  four  to  two 
against  him? 

21.  Two  gamblers  A  and  B  are  playing  a  game  of  chance  and  each 
player  has  staked  $32.  They  are  playing  for  three  points,  but  when  A  has 
gained  two  points  and  B  one  they  decide  to  stop  playing.  How  should 
they  divide  the  $64  ? 


424  A   COLLEGE   ALGEBRA 


XXVIII.     MATHEMATICAL   INDUCTION 

791  Mathematical  induction.  A  number  of  the  formulas  con- 
tained in  recent  chapters  may  be  established  by  a  method  of 
proof  called  mathematical  induction.  It  is  illustrated  in  the 
following  example. 

Example.     Prove  that  the  sum  of  the  first  n  odd  numbers  is  n^. 
We  are  asked  to  show  that 

l  +  3  +  5  +  ...  +  (2n-l)  =  n2.  (1) 

We  see  by  inspection  that  (1)  is  true  for  certain  values  of  n,  as  1  or  2. 
Suppose  that  we  have  thus  found  it  true  when  n  has  the  particular  value 
ki  so  that 

l  +  3  +  5  +  .--  +  (2A;-l)  =  A;2  (2) 

is  known  to  be  true.  Adding  the  next  odd  number,  naively,  2  (A;  +  1)  —  1, 
or  2  A;  +  1,  to  both  members  of  (2)  and  replacing  A;2  +  2  i  +  1  by  (A:  +  1)2, 
we  obtain 

l  +  3  +  5  +  ...  +  (2A;  +  l)  =  (A;  +  l)2.  (3) 

But  (3)  is  what  we  get  if  in  (1)  we  replace  n  by  A:  +  1.  We  have  there- 
fore .shown  that  if  (1)  is  true  when  n  has  any  particular  value  k,  it  is  also 
true  when  n  has  the  next  greater  value  k  +  \. 

But  we  have  already  found  by  inspection  that  (1)  is  true  when  k  has 
the  particular  value  1.  Hence  it  is  true  when  n  =  l  +  l,  or  2 ;  hence 
when  n  =  2  -I-  1,  or  3  ;  and  so  on  through  all  positive  integral  values  of  n, 
which  is  what  we  were  asked  to  demonstrate. 

And,  in  general,  if  a  formula  involving  n  has  been  found 
true  for  n  —  1  and  we  can  demonstrate  that  if  true  for  n  =  /.• 
it  is  also  true  for  w  =  Ar  +  1,  we  may  conclude  that  it  is  true 
for  all  positive  integral  values  of  n.  For  we  may  reason: 
Since  it  is  true  when  n  —  1,  it  is  also  true  when  n  =  1  +  1, 
or  2 ;  hence  when  n  =  2  +  l,ov  3;  and  so  on  through  all  posi- 
tive integral  values  of  n. 

As  another  illustration  of  this  method  we  add  the  following 
proof  of  the  binomial  theorem. 

For  small  values  of  n  we  find  by  actual  multiplication  that 

(a  +  6)"  =  a"  +  C'{a"-^b  +  C?^a"-^b^  +  •■•  +  0";.^-^  +  ■•■.      (1) 


THEORY   OF    EQUATIONS  426 

Multiplying  both  members  of  (1)  by  a  +  6,  we  obtain,  §  773,  1, 

(a  +  b)"  +  ^  =  a«  +  i  +  C'l\a"b+  C'^\a"-^b- f-  C;:     \a''-'-  +  ^b'-  +  •■- 

+  1  I      +^1  +c,ii\ 

+  C"  +  ' a(" +  !)-'■  6'-+ •••.  (2) 

But  (2)  is  the  same  as  (1)  with  n  replaced  by  n  +  I. 

Hence  if  (1)  is  true  when  n  =  fc,  it  is  also  true  when  n  =  k  +  1.  But 
(1)  is  known  to  be  true  when  n  =  1.  It  is  therefore  true  when  n  =  1  +  1, 
or  2  ;  therefore  when  n  =  2  +  1,  or  3 ;  and  so  on. 

Since  the  formula  C"  +  C,."i  =  C"  +  ^  can  be  proved  independently  of 
the  doctrine  of  combinations,  §  774,  the  proof  of  the  binomial  theorem 
nere  given  is  independent  of  that  doctrine. 

EXERCISE   LXIX 

Prove  the  truth  of  the  following  formulas,  §§  701,  712,  by  the  method  of 
mathematical  induction. 

1.  a  +  ar  +  ar'^  +  ■  ■  ■  +  ar" -^  =  a  (l  -  r")  / (1  -  r). 

2.  12  +  22  +  32  +  .  .  .  +  Ti2  =  ,1  (,i  4.  1)  (2  11  +  1)  /G. 

3.  13  +  23  +  33  + ..  •  +  n3  =  n2(n  +  l)-/4- 

4.  1  +  3  +  6  +  ••■  +  n(n  +  l)/2!  =  n{n  +  l)(n  +  2)/3!. 


XXIX.     THEORY    OF   EQUATIONS 

THE   FUNDAMENTAL  THEOREM.     RATIONAL    ROOTS 

The  two  standard  forms  of  the   general  equation  of  the  nth     792 
degree  in  x.     Every  rational  integral  equation  involving  a  single 
unknown  letter,  as  x,  and  of  the  nth  degree  with  respect  to 
that  letter,  can  be  reduced  to  the  standard  form 

a„x"  +  aiX"~^  +  •  •  •  +  (^n-i^  +  *«  =  0.  (1) 

When  the  coefficients  a^,  a^,  ■■■,  a„  are  given  numbers,  (1)  is 

called  a  numerical  equation,  but  when  they  are  left  wholly 

undetermined,  (1)  is  called  the  general  equation  of  the  wth 

degree. 


426  A   COLLEGE    ALGEBRA 

The  final  coefficient  a.,,  is  often  called  the  absolute  term. 

We  call  an  equation  of  the  form  (1)  complete  or  incomplete 
according  as  none  or  some  of  the  coefficients  aj,  02>  •  •  •?  «„  are  0. 
Observe  that  in  a  complete  equation  the  number  of  the  terms 
is  n  +  1. 

In  what  follows,  when  all  the  coefficients  a,,,  a^,  ■•-,  a„  are 
real  numbers,  we  may  and  shall  suppose  that  the  leading  one 
ttQ  is  positive,  and  when  they  are  rational,  that  they  are  integers 
which  have  no  common  factor. 

By  dividing  both  members  of  (1)  by  a^,  we  reduce  it  to  the 
second  standard  form 

x"  +  b^x"-^-] \-K-i^  +  K  =  Q,  (2) 

in  which  the  leading  coefficient  is  1,  and  b^  =  Oi/oq,  and  so  on. 
For  many  purjjoses  (2)  is  the  more  convenient  form  of  the 
equation. 

In  the  present  chapter  it  is  to  be  understood  that  f(x)  =  0 
denotes  an  equation  of  the  form  (1)  or  (2). 

Roots  of  equations.  The  roots  of  the  equation  f(x)  =  0  are 
the  values  of  x  for  which  the  polynomial  f(x)  vanishes, 
§§  332,  333.  It  is  sometimes  convenient  to  call  the  roots  of 
the  equation  the  roots  of  the  polynomial. 

From  the  definition  of  root  it  follows  that  when  o„  is  0 
one  of  the  roots  of /(a-)  =  0  is  0  ;  also  that  an  equation /(a-)  =  0 
all  of  whose  coefficients  are  positive  can  have  no  positive 
root,  and  that  a  complete  equation  f(x)  =  0  whose  coefficients 
are  alternately  positive  and  negative  can  have  no  negative 
root. 

Thu.s,  2x3  +  x^  +  1  =  0  can  have  no  positive  root  since  the  polynomial 
2  «*  +  x2  +  1  cannot  vanish  when  x  is  positive ;  and  2x^  — x^  +  Sx  —  1  =  0 
can  have  no  negative  root  since  2a;3  -  a;2  +  33.  _  i  cannot  vanish  when 

X  is  negative. 

Theorem  1 .  If  h  is  a  root  of  f  (x)  =  0,  then  f  (x)  is  exactly 
divisible  by  x  —  b;  and  conversely,  if  f{x)  is  exactly  divisible 
byx  —  h,  then  b  is  a  root  of  f  (x)  =  0. 


THEORY    OF    EQUATIONS  427 

For,  by  §  413,  the  remainder  in  the  division  of  /(a:)  by 
X  -b  is  f{b).  But  when  i  is  a  root  of  f{x)  =  0  this  remainder 
f{b)  is  0,  §  793,  so  that  f(x)  is  exactly  divisible  by  x  -b; 
and  conversely,  when  f(x)  is  exactly  divisible  by  x  -  b,  the 
remainder  f{b)  is  0,  so  that  b  is  &,  root  of  f(x)  =  0. 

Example.     Prove  that  3  is  a  root  of /(a;)  =  x^  -  2  x^  -  9  =  0. 
1     -2     +0     -9[3  Dividing  x^  -  2x2  _  9  ^y  x  -  3  synthet- 

3         3         9  ically,  §  411,  we  find  that  the  remainder /(3) 

1         1         3,        0=/(3)     is  0.     Hence  3  is  a  root  of /(x)  =  0. 

If  6  is  a  root  of  f(x)  =  0,  so  that  f(x)  is  exactly  divisible    796 
\)y  X  —  b,  and  we  call  the  quotient  cf>  (x),  we  have 
f(x)  =  (x-b)<t.(x). 

Hence  the  remaining  roots  of  f(x)  =  0  are  the  values  of  x 
for  which  the  polynomial  <f>  (x)  vanishes  ;  in  other  words,  they 
are  the  roots  of  the  depressed  equation  <fi  (x)  =  0,  §  341. 

Example.     Solve  the  equation  x^  -  3  x2  +  5  x  -  3  =  0. 
1     _  3     4-5     _  3  1 1  We  see  by  inspection  that  1  is  a  root,  and  divid- 

1     _  2     +  3  ~     ing  x3  -  3x2  +  5x  -  3  by  X  -  1,  we  obtain  the 
1     Z2     +^,        0  depressed  equation  x2  -  2  x  +  3  =  0.     The  roots 

of  this  quadratic,  found  by  §  631,  are  1  ±  i  V2.     Hence  the  roots  of  the 
given  equation  are  1,  1  +  i  V2,  and  1  —  i  v2. 

We  shall  assume  now  and  demonstrate  later  that  everi/     797 
rational  integral  equation  f(x)=  0  has  at  least  one  root. 

From  this  assumption  and  §  795  we  deduce  the  following 
theorem,  often  called  the  fundamental  theorem  of  algebra. 

Theorem  2.    Every  equation  of  the  nth  degree,  as  798 

f  (x)  =  aox"  +  aix°-i  +  •  •  •  +  a„_ix  +  a„  =  0, 
has  n  and  but  n  roots. 

By  §  797  there  is  a  value  of  x  for  which  f{x)  vanishes. 
Call  it  /81.  Then  f{x)  is  exactly  divisible  by  x  -  ^j,  §  795, 
the  leading  term  of  the  quotient  being  a^x"-'^.     Hence 

f{x)  =  {x- ft,)  {a.x'^-' +  ■■■).  (1) 


428  A   COLLEGE   ALGEBRA 

By  the  same  reasoning,  since  there  is  a  value  of  x,  call  it 

^2,  for  which  the  polynomial  Oo^"~^  H vanishes,  we  have 

a^"-^  -\ =  (a-  -  /3o)  (<'u.r"-2  ^ ).     Therefore,  by  (1), 

f{x)  =  {X  -  ^0  (X  -  ;8,)  (a,x'^-'  +  ■••).  (2) 

Continuing  thus,  after  n  divisions  we  obtain 

f{x)  =  a, {X  -  (30  {X-  p,)---{x-  ySJ.  (3) 

We  have  thus  shown  that  n  factors  of  the  first  degree  exist, 
namely,  x  -  ji^,x  -  ji^,- ■  ^x  -  j3„,  of  which /(a-)  is  the  product ; 
and  by  §419, /(x)  can  have  no  other  factors  than  these  and 
their  products. 

But  since  a  product  vanishes  when  one  of  its  factors  vanishes 
and  then  only,  it  follows  from  (3)  that  f(x)  vanishes  when 
X  =  /3i,  or  p.,  ■  ■■,  or  )8„,  and  then  only.  Hence,  §  793,  the  n 
numbers  /3i,  Pi,  ■  •■,  /8,,  are  roots  of  the  equation  f{x)  =  0  and  it 
has  no  other  roots  than  these. 

799  From  this  theorem  it  follows  that  the  problem  of  solving  an 
equation  f(x)  =  0  is  essentially  the  same  as  that  of  factoring 
the  polynomial  f(x).  Also  that  to  form  an  equation  which 
shall  have  certain  given  numbers  for  its  roots,  we  have  merely 
to  subtract  each  of  these  numbers  in  turn  from  x,  and  then  to 
equate  to  0  the  product  of  the  binomial  factors  thus  obtained. 

Example.     Form  the  equation  whose  roots  are  2,  1/2,  -  1,  0. 

It  i.s  (X  -  2)  (X  -  1/2)  (X  +  1)  (x  -  0)  =  0,  or  2x4  -  3x5  -  3x2  +  2x  =  q. 

800  Multiple  roots.  Observe  that  two  or  more  of  the  roots  ^i, 
Pi,  ■■■,  /8„  may  be  equal.  If  two  or  more  of  them  are  equal  to 
P,  we  call  p  a  viuJtiple  root.  And  according  as  the  number  of 
the  roots  equal  to  /8  is  two,  three,  ■  ■•,  in  general,  r,  we  call  p  a 
double  root,  a  triple  root,  in  general,  a  root  of  order  r.  A  simple 
root  may  be  described  as  a  root  whose  order  r  is  1.  Evidently 
it  follows  from  §  798  that 

The  condition  that  p  he  a  root  of  order  r  "/  f  (x)  =  0  is  that 
f  (x)  he  divisible  by  (x  —  py  but  not  by  (x  —  /8)'  +  ^ 


THEORY    OF   EQUATIONS  429 

When  we  say,  therefore,  that  every  equation  of  the  nth 
degree  has  n  roots,  the  understanding  is  that  each  multiple 
root  of  order  r  is  to  be  counted  r  times.  It  is  of  course  7iot 
true  that  every  equation  of  the  ?ith  degree  has  n  different 
roots. 

Thus,  x^  —  3x2  +  3  a;  —  1  =  0  is  an  equation  of  the  third  degree ;  but 
since  x^  —  3x2-f3x  —  l  =  (x  —  l)^,  each  of  its  roots  is  X. 

On   finding   the   rational   roots    of   numerical   equations.     Let     801 

f{x)  =  a^pc"  +  Oja;""'  +  •  •  •  +  «„  =  0  denote  an  equation  with 
integral  coefficients,  and  let  b  denote  an  integer  and  b  / c  a, 
rational  fraction  in  its  lowest  terms.  It  follows  from  §§  451, 
795  that  if  h  is  a  root  of  f{x)  —  0,  then  i  is  a  factor  of  a„ ; 
and  from  §§  452,  795  that  if  Zi/e  is  a  root,  then  J  is  a  factor 
of  a„  and  c  is  a  factor  of  a^.  Hence,  in  particular,  if  a^  =  1> 
b  /c  cannot  be  a  root  unless  c  =  ±  1,  that  is,  unless  b  /c  denotes 
the  integer  ±  b.     Hence  the  following  theorem,  §  454  : 

An  equation  of  the  form  x"  +  ajx""'  -f  . . .  -f  a„  =  0,  cohere 
ai,  •••,  a„  denote  integers,  cannot  have  a  rational  fractional 
root. 

It  follows  from  what  has  just  been  said  that  all  the  rational     803 
roots  of  an  equation  with  rational  coefficients  can  be  found 
by  a  limited  number  of  tests.     These  tests  are  readily  made  by 
synthetic  division. 

Example.     Find  the  rational  roots,  if  any,  of  the  equation 

3  x5  -  8  X*  +  x2  +  12  X  +  4  =  0. 
The  only  possible  rational  roots  are  ±1,  ±2,  ±4,  ±1/3,  ±  2/3,  ±  4/3. 
We  see  by  inspection  that  1  is 
not  a  root.     Testing  2,  we  find  that 
it  is  a  root  and  obtain  the  depressed 
equation  3x^-2 x3-4x2-7x-2=0. 
We  find  that  2  is  a  root  of  this 
depressed    equation    also    and    ob- 
tain the  second  depressed  equation 
I  x^  +  4  x2  +  4  X  +  1  =  0.     This  equation  can  have  no  positive  root  since 


-8 

6 

-2 

+  0 
-4 
-4 

+  1 
-8 

-7 

+  12 

-  14 

-  2, 

+  4[2 
-  4 

0(2 

6 
4 

8 
4 

8 
1, 

2 

-\/?> 

-1 

-  1 

-1 

3 

3, 

0 

430  A   COLLEGE   ALGEBRA 

all  its  terms  are  positive,  §  794.     Testing  —1,  we  find  that  it  is  not  a  root  . 
Testing  —  1/3,  we  find  that  it  is  a  root  and  obtain  the  third  depressed 
equation  x^  +  x  +  1  —  0. 

Hence  the  rational  roots  of  the  given  equation  are  2,  2,  —\/Z.  Its 
remaining  roots,  found  by  solving  x^  +  x  +  1  =  0,  are  (—  1  ±  i  V3)/2. 

803  The  reckoning  involved  in  making  these  tests  will  be  lessened 
if  one  bears  in  mind  the  remark  made  in  §  453 ;  also  the  fact 
that  a  number  known  not  to  be  a  root  of  the  given  equation 
cannot  be  a  root  of  one  of  the  depressed  equations,  §  796 ;  and 
finally  the  following  theorem  : 

If  b  is  2}ositive  and  the  signs  of  all  the  coefficients  in  the 
result  of  dividing  f  (x)  byx.  —  b  synthetically  are  plus,  f  (x)  =  0 
can  have  no  root  greater  than  b  ;  ifh  is  negative  and  the  signs 
just  mentioned  are  alternately  2)lus  and  rninus,  f(x)  =  0  can 
have  710  root  algebraically  less  tlian  b. 

Eor  it  follows  from  the  nature  of  synthetic  division  that  in 
both  cases  the  effect  of  increasing  h  numerically  will  be  to 
increase  the  numerical  values  of  all  coefficients  after  the  first 
in  the  result  without  changing  their  signs,  so  that  the  final 
coefficient,  that  is,  the  remainder,  cannot  be  0. 

Example  1.  Show  that  2x3+3x2-4x4-5  =  0  has  no  root  greater  than  1. 

2  +3     — 4  +5|J.  Dividing  by  x  —  1,  we  obtain  positive  coefli- 

_     2     5  1  cients   only.       Hence   there   is  no   root  gi'eater 

2+5+1,        6         than  1. 

If  we  divide  by  x  —  2,  we  obtain  a  result  with  larger  coefficients,  all 
positive,  namely,  2  +  7  +  10,  25. 

Example  2.    Showthat  3x3+4x2-3x  +  l  =  0  has  no  root  less  than  -2. 

3  +4  -  3  +1  I  -  2  Dividing  by  x  +  2,  we  obtain  coefficients 
_  ~  ^  +  4  —  2  which  are  alternately  plus  and  minus.  Hence 
3     —  2     +1,-1              tliere  is  no  root  less  than  —  2. 

If  we  divide  by  x  +  3,  we  obtain  coefficients  with  the  same  signs  as  those 
just  found  but  numerically  greater,  namely,  3-5  +  12,  -  35. 

804  We  may  add  that  any  number  which  is  known  to  be  alge- 
braically greater  than  all  the  real  roots  of  f{x)  =  0  is  called  a 


THEORY   OF    EQUATIONS  431 

sujjerior  limit  of  these  roots,  and  that  any  number  which  is 
known  to  be  algebraically  less  than  all  the  real  roots  oi  f(x)  —  0 
is  called  an  inferior  limit  of  these  roots. 

Thus,  we  have  just  proved  that  1  is  a  superior  limit  of  the  roots  of 
2x^  +  3x2  —  4x  +  5  =  0  and  that  —  2  is  an  inferior  limit  of  the  roots 
of  3x3  +  4x2-3x  +  l  =^0. 

EXERCISE   LXX 

1.  Form  the  equations  whose  roots  are 

(1)  a,  -  6,  a  +  6.  (2)  3,  4,  1/2,  -  1/3,  0. 

2.  Show  that  —  3  is  a  triple  root  of  the  equation 

X*  +  Sx''  +  18  X-  -  27  =  0. 

3.  Show  that  1  and  1/2  are  double  roots  of  the  equation 

4  x5  -  23  x5  +  33  x2  -  17  X  +  3  =  0. 

4.  By  the  method  of  §  803  find  superior  and  inferior  limits  of  the  real 
roots  of  x5  —  Sx*  —  5x^  +  4  x^  —  7  X  —  250  =  0. 

5.  Show  that  2x*  —  3x^  +  4x2  —  lOx  —  3  =  0  has  no  rational  root. 

Each  of  the  following  equations  has  one  or  more  rational  roots.  Solve 
them. 

6.  x3  -  x2  -  14  X  +  24  =  0.  7.    x^  -  2  x2  -  25  x  +  50  =  0. 
8.    3  x^  -  2  x2  +  2  X  +  1  =  0.  9.    2  x<  +  7  x^  -  2  x^  -  x  =  0. 

10.  X*  4-  4  x3  +  8  x2  +  8  X  +  3  =  0. 

11.  2  X*  +  7  x3  +  4  x2  -  7  X  -  6  =  0. 

12.  3x1 +  11x3  + 9x2 +  11X  + 6  =  0. 

13.  x5  -  9  X*  +  2  x3  +  71  x2  +  81  X  +  70  =  0. 

14.  2x5 -8x*  +  7x3  +  5x2-8x  +  4  =  0. 

15.  x5 +  3x* -15x3- 35x2  + 54x+ 72  =  0. 

16.  12x4  -  32x3  +  13x2  +  8x  -  4  =  0. 

17.  x5  -  7x4  +  10x3  +  18x2 -27x- 27=0. 

18.  2x*  -  17x3  +  25x2 +  74x- 120  =  0. 


I  432  A  COLLEGE  ALGEBRA 

19.  4  x»  -  9  x-^  +  6  x2  -  13  X  +  0  =  0. 

20.  x5  +  8x-«  +  3x3  -  80x2  _  52x  +  240  =  0. 

21.  2  x5  +  11  X*  +  23  x-  +  25  x2  +  10  X  +  4  =  0. 

22.  6x*  -  89x5  +  359x2  _  254x  +  48  =  0. 

23.  lOx*  +  41x'' +  46x2 +  20X  + 3  =  0. 

24.  30 X*  -  108x3  +  107  x2  -  43x  +  6  =  0. 

25.  12x5  +  20x*  +  29x3  +  77x2  +  COx  +  18  =  0. 

26.  2  x6  +  7  x5  +  8  X*  +  7  x3  +  2  x2  -  14  x  -  12  =  0. 

27.  2x6  +  11x5  _^  24x*  +  22x3- 8x2 -.33x- 18  =  0. 

28.  5  x6  -  7  x5  -  8  X*  -  x3  +  7  x2  +  8  X  -  4  =  0. 

RELATIONS    BETWEEN   ROOTS   AND   COEFFICIENTS 

805  Relations  between  roots  and  coefficients.  When  an  equation 
whose  roots  are  (3i,  fS-i,  •  •  •?  y8„  is  reduced  to  the  second  standard 
form,  §  792,  (2),  the  identity  in  §  798,  (3),  becomes 

X"  +  b^x"-^  +  h„x"--  +  h^x"-^  H \-b„ 

=  (x  -  /30  (,r  -  p,)  {X  -  /3.)  • .  •  {X  -  p,). 

Carry  out  the  multiplications  indicated  in  the  second  mem- 
ber and  arrange  the  result  as  a  polynomial  in  x,  §  559.  Then 
equate  the  coefficients  of  like  powers  of  x  in  the  two  members, 
§  284.  We  thus  obtain  the  following  relations  between  the 
coefficients  h^^,  h,,,  ■■■,  ^>„  and  the  roots  /?i,  /?o,  •••,  y8„ : 

-^  =  i8i  +  /8, +  i8, +  ---  +  )8,„  (1) 

b,  =  p,/3,  +  13,(3,  +  •  •  •  +  A/?3  +  •  •  •  +  f3,.^^/3,„       (2) 

-  ^3  =  AA/Sa  +  A/8.i8,  +  •  •  •  +  /3,_, A,-i/8,„  (3) 


(-iyb„  =  p,p,p,...l3,„  ^n) 

where  the  second  members  of  (2),  (3),  ■••  represent  the  sum  of 
the  products  of  every  two  of  the  roots,  of  every  three,  and  so  on, 


THEORY    OF    EQUATIONS  433 

and  the  sign  before  the  first  member  is  plus  or  minus  according 
as  the  number  of  the  roots  in  each  term  of  the  second  member 
is  even  or  odd.     Hence  the  theorem  : 

Theorem.     In  every  equation  reduced  to  the  form  806 

X"  +  bix"-'  +  Kx"-2  +  . . .  +  b„  =  0, 

tlie  coefficient  bj  of  the  second  term,  ivith  its  sign  changed,  is 
equal  to  the  sum  of  all  the  roots;  the  absolute  term  b„,  urith  its 
sign  changed  or  not  according  as  n  is  odd  or  even,  is  equal  to 
til  e  product  of  all  the  roots  ;  and  the  coefficient  \  of  each  inter- 
mediate term,  with  its  sign  changed  or  not  according  as  r  is 
odd  or  even,  is  equal  to  the  sum  of  the  products  of  every  r  of 
the  roots. 

Before  applying  this  theorem  to  an  equation  whose  lead- 
ing coefficient  is  not  1  we  must  divide  the  equation  by  that 
coefficient.  If  the  equation  be  incomplete,  it  must  be  remem- 
bered that  the  coefficients  of  the  missing  terms  are  0. 

Thus,  without  solving  the  equation  Sx-'^  —  Gx  +  2  =  0,  we  know  the 
following  facts  regarding  its  roots  /3i,  p«,  (is.  Reduced  to  the  proper 
form  for  applying  the  theorem,  the  eciuation  is  x^  +  Ox^  —  2  x  +  2/3  =  0. 
HcncG  ' 

/3i  +  /32  +  /33  =  0,  /3i/32  +  /3i/33  +  ^2^3  =  -  2,  ^i^o/Ss  =  -  2 /3. 

If  all  but  one  of  the  roots  of  an  equation  are  known,  we  can     807 
find  the  remaining  root  by  subtracting  the  sum  of  the  known 
roots  from  —  bi,  or  by  dividing  b,„  with  its  sign  changed  if  71  is 
odd,  by  the  product  of  the  known  roots. 

Example.  Two  of  the  roots  of  2  x^  +  3  x^  -  23  x  -  12  =  0  are  3  and 
—  4.     What  is  the  remaining  root  ? 

The  remaining  root  is  -  3/2  -  [3  +  (-  4)]  =  -  1/2  ;  or  again,  it  is 
6-3(-4)=-l/2. 

When  the  roots  themselves  are  connected  by  some  given     808 
relation,  a  corresponding  relation  must  exist  among  the  coeffi- 
cients.    To  find  this  relation  we  apply  the  theorem  of  §  806. 


434  A   COLLEGE   ALGEBRA 

Example  1.     Find  the  condition  that  the  roots  of  x^  +  px^  ^qx  +  r  =  0 
shall  be  in  geometrical  progression. 

Representing  the  roots  by  a  / p,  a,  a^,  we  have 

-  +  a  +  a/3  =  -p,  —  +  a2  _f,  ^2^  =  g^  _.  a  ■  a^  =  -r. 
/3  ^  ^ 


The  third  equation  reduces  to  a^  =  —  r,  whence  a  =\  —  r. 
Dividing  the  second  equation  by  the  first,  substituting  a  =  V—  r  in  the 
result,  and  simplifying,  we  have  q^  —  p^r  =  0. 

Example  2.     Solve  the  equation  x^  +  8  x^  4-  5  x  —  50  =  0,  having  given 
that  it  has  a  double  root. 

Representing  the  roots  by  a,  a,  /3,  we  have 

2  a  +  ^  =  -  8,   a:2  +  2  a/3  =  5,   a2/3  =  60. 

Solving  the  first  and  second  of  these  equations  for  a  and  /3,  we  obtain 
a  =  -5,  ^  =  2  and  a  =  -1/3,  ^  =  -22/3. 

The  values  a  =  —  5,  /3  =  2  satisfy  the  equation  a^j3  =  50,  but  the  values 
a  =  —  1/3,  ^  =  —  22/3  do  not  satisfy  this  equation. 

Hence  the  required  roots  are  —  5,  —  5,  2. 

809  Symmetric  functions  of  the  roots.  The  expressions  in  the 
roots  to  which  the  several  coefficients  are  equal,  §  805,  are  si/m- 
metric  functions  of  the  roots,  §  540.  It  will  be  proved  in  §  868 
that  all  other  rational  symmetric  functions  of  the  roots  can  be 
expressed  rationally  in  terms  of  these  functions,  and  therefore 
rationally  in  terms  of  the  coefficients  of  the  equation. 

Example  1.     Find  the  sum  of  the  squares  of  the  roots  of  the  equation 
2x3-3x2-4x-  5  =  0. 

Calling  the  roots  a,  /3,  7,  we  have 

a2  +  ^2  +  72  =  (a  +  /3  +  7)2  -  2 (a/3  +  ^7  +  7«)  =  (3/2)2  +  4=61. 

Example  2.     If  the  roots  of  x'  +  px"^  +  qx  +  r  =  0  are  a,  /3,  7,  what 
is  the  equation  whose  roots  are  ^y,  ya,  a/3  ? 

If  p',  g',  r'  denote  the  coeflHcients  of  the  required  equation,  we  have 

-  p'  =  /37  +  7a  +  a/3  =  7, 

g'  =  ^7  .  7a  +  7a  •  a/3  +  a/3  ■  ^y 

=  a/37  (or  +  |3  +  7)  =  ( -  r)  ( -  i))  =  rp, 

-  r'  =  /37  •  7a  •  a/3  =  (a^7)2  =  r"^. 
Hence  the  required  equation  is  x^  —  qx^  +  prx  —  r2  =  0. 


THEORY    OF    EQUATIONS  435 


EXERCISE  LXXI 

1.  Two  of  the  roots  of  2  x^  -  7  a;^  +  lo  j  -  6  =  0  are  1  ±  i ;  find  the 
third  root. 

2.  The  roots  of  each  of  the  following  equations  are  in  geometrical 
progression ;  find  them. 

(1)  8x3-14x2-21x4-27  =  0.      (2)  x^  +  x2  +  3x  +  27  =  0. 

3.  The  roots  of  each  of  the  following  equations  are  in  arithmetical 
progression ;  find  them. 

(1)  x3  +  6x2  +  7x-2  =  0.  (2)  x3  _  9x2 +  23x- 15  =  0. 

4.  Show  that  if  one  root  of  x^  +  px2  +  gx  +  r  =  0  be  the  negative  of 
another  root,  pq  —  r. 

5.  Find  the  condition  that  one  root  of  x^  +  px-  +  qx  +  r  =  0  shall  be 
the  reciprocal  of  another  root. 

6.  Solve  X*  +  4.x3  +  10x2  +  12x  +  9  =  0,  having  given  that  it  has 
two  double  roots. 

7.  Solve  the  equation  14x3  —  13x2  —  18x  +  9  =  0,  having  given  that 
its  roots  are  in  harmonical  progression. 

8.  Solve  the  equation  x*  —  x^  -  56  x2  +  36  x  +  720  =  0,  having  given 
that  two  of  its  roots  are  in  the  ratio  2  : 3  and  that  the  difference  between 
the  other  two  roots  is  1. 

9.  If  a,  i3,  7  are  the  roots  of  x^  +  px-  +  gx  +  r  =  0,  find  the  equa- 
tions whose  roots  are 

(1)   -  a,  -  i3,  -  7.  (2)  fca,  k^,  ky. 

(3)  1/a,  1//?,  1/7.  (4)  a  +  fc, /3  +  A:,  7  + A;. 

(5)  a^,  ^,  72.  (6)   -  l/a2,  _l//32,  -  1/72. 

10.  If  a,  i3,  7  are  the  roots  of  2  x^  +  x2  -  4  x  +  1  =  0,  find  the  values  of 

(1)  a2  +  ^2  +  y2.  (2)  a3  +  ;33  +  73. 

(3)  l/^y+l/ya+l/a^.         (4)  a/32  +  ^,^2  +  ^^2  +  ^^2  +  ^2,^  +  a^. 

11.  If  a,  /3,  7  are  the  roots  of  x^  -  2  x2  +  x  -  3  =  0,  find  the  values  of 
(1)  a/py  +  ^/ya  +  y/a^         (2)  a^/y  +  ^y/ a  +  ycx /^. 

(3)  (^  +  7)  (7  +  «)  («  +  ^)-         (4)  (/32  +  7^)  (72  +  a"')  {a^  +  ^). 


436  A   COLLEGE   ALGEBRA 

TRANSFORMATIONS   OF  EQUATIONS 

Some  important  transformations.  It  is  sometimes  advan 
tageous  to  transform  a  given  equation  f{x)  =  0  into  another 
equation  whose  roots  stand  in  some  given  relation  to  the  roots 
of  f(x)  =  0.  The  transformations  most  frequently  used  are 
the  following : 

To  transform  a  given  equation  f  (x)=  0  into  another  tvhose 
roots  are  those  o/f  (x)=  0  tvith  their  signs  changed. 

The  required  equation  is  /(—  y)  =  0.  For  substituting  any 
number,  as  y8,  for  x  in  f(x)  gives  the  same  result  as  substitut- 
ing —  /8  for  y  in  /(—  ij).  Hence,  if /(a;)  vanishes'when  x  =  /3, 
y(_  y)  will  vanish  when  y  =—  P;  that  is,  if  /S  is  a  root  of 
fix)  =  0,  -  ^  is  a  root  of  /(-  y)  =  0. 

Therefore  every  root  of  f(x)  =  0,  with  its  sign  changed, 
is  a  root  of  /(—  y)  =  0  ;  and  /(—?/)=  0  has  no  other  roots 
than  these,  since  f(x)  =  0  and  /(-?/)  =  0  are  of  the  same 
degree. 

If  the  given  equation  is 

rt^cc"  +  «ia;"~*  +  a^x"'^  +  •  •  •  -f-  a„  =  0, 
the  required  equation  will  be 

«.,(-  t/T  +  «i  (-  ?/)""'  +  a,{-  y)"-'  +  •  •  •  +  «^„  =  0, 

which  on  being  simplified  becomes 

«oy"  -  «iy""'  +  o^y"'- +  (-  !)"«« =  O- 

Hence  the  required  equation  may  be  obtained  from  the  given 
one  by  changing  the  signs  of  the  terms  of  odd  degree  when  n 
is  even,  and  by  changing  the  sigyis  of  the  terms  of  even  degree, 
including  the  absolute  term,  when  n  is  odd. 

We  may  use  x  instead  of  y  for  the  unknown  letter  in  the 
transformed  equation,  and  write /(—  a;)  =  0  for /(—  y)  =  0. 


THEORY    OF    EQUATIONS  437 

Example.     Change  the  signs  of  the  roots  of 

4x5  -  9x3  +  6x2  -  18x  +  6  =  q. 
Changing  the  signs  of  the  terms  of  even  degree,  we  have 
4x5  -  9x3  -  6x2  -  13x  -  6  =  o. 
In  fact,  the  roots  of  the  given  equation  are  1  / 2,  3/2,  —  2,  ±  i,  and  those 
of  the  transformed  equation  are  —  1/2,  —  3/2,  2,  T  i- 

To  transfortn  a  given  equation  f(x)=  0  into  another  ichose     812 
roots  are  those  o/f  (x)  —  0  eacli  multiplied  by  some  constant,  as  k. 

The  required  equation  is  f(i//k)  =  0.  For  if /(a*)  vanishes 
when  X  =  P,f{i//k)  will  vanish  when  y /k  =  p,  that  is,  when 
y  —  k(i  (compare  §  811). 

If  the  given  equation  is 

aoX"  +  aix"-^  +  a^x""-^  -\ 1-  a„  =  0, 

the  required  equation  will  be 

which  when  cleared  of  fractions  becomes 

«(,?/"  +  kaiy"~^  +  k'^Ooy"'^  +  •  •  •  +  k"a,^  =  0. 

Hence  the  required  equation  may  be  obtained  by  multiply- 
ing the  second  term  of  the  given  equation  by  k,  its  third  term 
by  k^,  and  so  on,  taking  account  of  missing  terms  if  any. 

When  k<=  —  l  this  transformation  reduces  to  that  of  §  811. 

Example.  Multiply  the  roots  of  x*  +  2x3  -  x  +  3  =  0  by  2.  Also 
divide  them  by  2. 

The  first  of  the  required  equations  is  x*  +  4  x3  —  8  x  +  48  =  0,  and  since 
dividing  by  2  is  the  same  as  multiplying  by  1/2,  the  second  is 

a;4  +  x3-x/8  +  3/lG  =  0,  or  lOx*  +  16x3  -  2  x  +  3  =  0. 

The  following  example  illustrates  an  important  application     813 
of  the  transformation  now  under  consideration. 

Example.  Transform  the  equation  36x3  +  18x2  +  2x  +  9  =  0  into 
another  whose  leading  coefficient  is  1  and  its  remaining  coefficients 
integers. 

Dividing  by  36,  we  have 

x3  +  x2/2  +  x/18  + 1/4  =  0.  (1) 


438  A  COLLEGE    ALGEBRA 

Multiplying  the  roots  by  k, 

x3  +  A-xV2  +  fc2x/18  +  fcV4  =  0.  (2) 

"We  see  by  inspection  that  the  smallest  value  of  k  which  will  cancel  all 
the  denominators  is  6.     And  substituting  6  for  k  in  (2),  we  have 

x3  +  3x2 +  2x  + 54=0,  (3) 

■which  has  the  form  required.     The  roots  of  (3)  each  divided  by  6  are  the 
roots  of  the  given  equation  (1). 

814  To  transform  a  f/ive?i  equation  f(x)  =  0  into  another  tvhose 
roots  are  the  reciprocals  of  those  of  f  (x)  =  0. 

The  required  equation  is /(I/?/)  =  0.     For  \i  f{x)  vanishes 
when  X  =  (3,  /(1/y)  will  vanish  when  1/y  =  {3,  that  is,  when 

If  the  given  equation  is 

CIqX"  +  (liX"~^  +  •  •  •   +  <^n-V^  +  ^n  =  0, 

the  required  equation  will  be 

«o/y'  +  «i/y"'  +  •  •  •  +  a„_i/y  +  «„  =  0, 

which  when  cleared  of  fractions  becomes 

ay  +  «„_i//'"^  H +  aiy  +  ^0  =  0. 

Hence  the  required  equation  may  be  obtained  by  merely 
reversing  the  order  of  the  coefficients  of  the  given  equation. 

Example.     Replace  the  roots  of  2  x*  -  x2  -  3  x  +  4  =  0  by  their  recip- 
rocals. 

Reversing  the  coefficients,  we  have  4x*  -  Sx^  —  x^  +  2  =  0. 

815  An  equation  like  2  x^  -\-  3  x^  —  Z  x  —  2  =  0,  which  remains 
unchanged  when  this  transformation  is  applied  to  it,  that 
is,  when  the  order  of  its  coefficients  is  reversed,  is  called  a 
reciprocal  equation,  §  645.  If  /3  is  a  root  of  such  an  equation, 
l/;8  must  also  be  a  root.  Hence  when  the  degree  of  the 
equation  is  even,  half  of  the  roots  are  the  reciprocals  of  the 
other  half.  The  same  is  true  of  all  the  roots  but  one  when 
the  degree  is  odd;  but  in  this  case  there  must  be  one  root 
which  is  its  own  reciprocal,  that  is,  one  root  which  is  either 


THEORY    OF    EQUATIONS  439 

1  or  -  1.     Thus,  one  root  of2a;^4-3x2-3cc-2  =  0isl  and 
the  otlier  roots  are  —  2  and  —1/2. 

From  the  nature  of  this  transformation  it  follows  that  ivhen     816 
an  equation  has  variable  coefficients  and  the  leading  coefficient 
vanishes,  one  of  the  roots  becomes  infiiiite;  when  the  tivo  leading 
coefficients  vanish,  two  of  the  roots  become  infinite;  and  so  on. 

Example  1.  Show  that  one  of  the  roots  of  mx^  +  3x2_2x  +  l=0 
becomes  infinite  when  m  vanishes. 

Applying  §  8U  to       mx^  +  3  x^  -  2  a;  +  1  =  0,  (1) 

we  obtain  x»  -  2x2  +  3x  +  m  =  0.  (2) 

If  the  roots  of  (2)  are  ^i,  /Sg,  /Sg,  those  of  (1)  are  l/^i,  l/iSj,  1//33. 

By  §  806,  i3i/32/33  =  —  m.  Hence,  if  m  approach  0  as  limit,  one  of  the 
roots  /3i,  /3o,  /33  must  also  approach  0  as  limit,  and  if  this  root  be  /3i,  the 
corresponding  root  of  (1),  namely,  1  /^i,  must  approach  oo,  §  512. 

Example  2.  Show  that  two  of  the  roots  of  mx^  +  iii^x-  +  x  +  1  =  0 
become  infinite  when  m  vanishes. 

Applying  §  814  to        mx^  +  m^x'^  +  x  +  1  ^  0,  (1) 

we  have  x^  +  x'^  +  ni^x  +  m  -  0.  (2) 

If  the  roots  of  (2)  are  ^i,  ^2,  Pz,  those  of  (1)  are  l//3i,  1//32,  I  / ^z- 

^1^3  =  -m,     /3ii32  +  ^1^3  +  ftft  = '«^-  (3) 

It  follows  from  (3)  that  if  m  approach  0,  two  of  the  three  roots  ^1,  ^2,  ^3 
must  also  approach  0,  and  if  these  roots  be  /3i,  /Sa,  the  corresponding  roots 
of  (1),  namely,  l//3i,  1//32,  must  approach  oo. 

To  transform  a  given  eqxiation  f  (x)=  0  into  another  whose     817 
roots  are  those  of  f  (x)  =  0,  each  diminished  by  some  constant, 
as  k. 

The  required  equation  is/(y  +  ^)  =  0.  For  \ff{x)  vanishes 
when  x  =  /3,  f{y  +  k)  will  vanish  when  y  +  k  =  jS,  that  is, 
when  y  =  /3  —  k. 

If  the  given  equation  is 

f(x)  =  a^x"  +  Oiic"  -'  + [-  a„_^x  +  a„  =  0. 

the  required  equation  will  be 

f(y  +  k)  =  a,(y  +  ky  +  a,(y  +  k)"-'  +  . . .  +  a„  =  0, 


440  A   COLLEGE   ALGEBRA 

which,  when  its  terms  are  expanded  by  the  binomial  theorem 
and  then  collected,  will  reduce  to  the  form 

where  Cq  =  «„,  Ci  =  nkao  +  ai,  and  so  on. 

This  method  of  obtaining  <^  (y)  from  f(x)  is  usually  very 
laborious.  The  following  method  is  much  more  expeditious,  at 
least  when  the  coefficients  otf(x)  are  given  rational  numbers. 

It  X  =  i/  +  k,  then  y  =  x  —  k,  and  we  have 

/(^)=/(y  +  k)=<t>(y)=<l>(x  -  k), 
that  is, 

Co(x  —  k)"  -\ h  c„_i  (x  —  k)  +  c„  =  a^"  H \- a„_iX  +  a„. 

If  both  members  of  this  identity  be  divided  by  x  —  k,  if 
again  the  quotients  thus  obtained  be  divided  by  x  —  k,  and 
so  on,  the  successive  remainders  yielded  by  the  first  member, 
namely,  c„,  c„_,,  •  •  •,  will  be  the  same  as  those  yielded  by  the 
second  member.  Hence  we  may  obtain  ^  (ij)  from  f{x)  as 
follows:  Divide  f  (x)  hij  x  —  k,  divide  the  quotient  thus  obtained 
hij  X  —  k,  and  so  on.  The  successive  remainders  will  be  c„, 
Cn_i,  •••,  Ci,  and  the  final  quotient  will  be  %  (compare  §  423). 
The  divisions  should  be  performed  synthetically. 

Example  1.     Diminish  the  roots  of  2  a;3  -  7  x^  _  3x  +  1  =  0  by  4. 
First  method.     Substituting  ?/  +  4  for  x,  we  have 

2  x3  _  7  x2  _  3  X  +  1  =  2  (y  +  4)3  -  7  (y  +  4)2  -  3  (y  +  4)  +  1 
=  2  2/3 +  17  2/2 +  37  2/ +  5. 
Second  method.     Arranging  the  reckoning  as  in  §  423,  we  have 

2     -    7     -    3     +1  [4 

8     4     _4 

2+1+1,        5        .-.03  =  5. 

8     _36 

2     +    9,        37  .-.  C2  =  37. 

8 

2,        17  .-.  cx^  17  and  Co  =  2, 

Hence,  as  before,  we  find  the  required  equation  to  be 
2  2/3  +  17  y2  +  37  y  +  5  =  0. 


THEORY    OF   EQUATIONS  441 

Example  2.     Increase  the  roots  of  x^  +  4  x^  +  x  +  3  =:  0  by  4. 

To  increase  the  roots  by  4  is  the  same  thing  as  to  diminish  them  by 

-  4.  Hence  the  required  equation  may  be  obtained  either  by  substitut- 
ing 2/  —  4  for  X  or  by  dividing  synthetically  by  —  4.  It  will  be  found  to 
be  2/3  -  8  2/2  +  17  2/  -  1  =  0. 

By  aid  of  §  817  we  can  transform  a  given  equation  into     818 
another  which  lacks  some  particular  power  of  the  unknown 
letter. 

Example  1.  Transform  the  equation  x^  — 3x2  +  5x  +  6  =  0  into 
another  which  lacks  the  second  power  of  the  unknown  letter. 

Substituting  x  =  y  +  k,  we  have  y'^  +  {Sk  —  3)y'^  +  •  ■  ■ .  Hence  we 
must  have  3  A:  — 3  =  0,  that  is,  fc  =  1.  And  diminishing  the  roots  of 
x3  -  3x2  +  5x  +  6  =  0  by  i^  ^^e  obtain  x^  +  2x  +  9  =  0. 

Example  2.  Transform  the  equation  x*^  —  5  x'-  +  8  x  —  1  =  0  into 
another  which  lacks  the  first  power  of  the  unknown  letter. 

Substituting  x  =  y  +  k,  we  have 

2/3  +  (3  /^  _  5)  2/2  +  (3  ^2  _  10  A:  +  8)  2/  +  •  •  •  =  0. 
Hence  we  must  have  3  fc^  _  lO  A:  +  8  =  0,  that  is.  A;  =  2  or  4/3. 
Diminishing  the  roots  of  x^  —  5  X'^  +  8  x  —  1  =  0  by  2,  we  obtain 
x3  +  x2  +  3  =  0. 

If,  when  we  diminish  the  roots  of  f(x)  =  0  hy  k,  we  obtain  819 
an  equation  ^(a:r)=0  whose  terms  are  all  positive,  ^  is  a 
superior  limit  of  the  positive  roots  of /(a')=  0,  §  804.  For  in 
this  case  (f>(x)=0  has  no  positive  root,  §794.  Hence  any- 
positive  roots  that  f(x)  =  0  may  have  become  negative  when 
diminished  by  k.     They  are  therefore  less  than  k. 

The  process  of  synthetic  division  is  such  that  it  is  possible 
by  inspection  and  trial  to  find  the  smallest  integer  k  for  which 
all  the  terms  of  <fi(x)=  0  will  be  positive.  I'n  most  cases  this 
can  be  accomplished  with  comparatively  little  labor. 

We  may  obtain  an  inferior  limit  of  the  negative  roots  of 
f(x)=  0  by  finding  a  superior  limit  of  the  roots  of /(—  x)  =  0. 
For  if  A;  is  a  superior  limit  of  the  roots  of  /(—  a;)=  0,  then 

—  A;  is  an  inferior  limit  of  the  roots  oi  f{x)  =  0,  §  811. 


442  A    COLLEGE    ALGEBR.A 

Example.  Find  superior  and  inferior  limits  of  the  roots  of  the  equa- 
tion /(x.)  =  x*  -  6  z3  +  14  z2  +  48  X  -  121  =  0. 

We  find  by  inspection  and  trial  that  neither  A;  =  lnorA;=32  will  give 
a  transformed  equation  <p(x)  =  0  all  of  whose  terms  are  positive,  but 
that  k  =  S  will.  In  fact,  if  we  diminish  the  roots  of  f{z)  =  0  by  3,  we 
obtain  0  (x)  =  x*  +  6  x^  +  14  x^  +  78  x  +  68  =  0.  Hence  3  is  a  superior 
limit  of  the  roots  of  /(x)  =  0. 

The  equation  /( -  x)  =  0  is  x*  +  6  x''  +  14  x^  -  48  x  -  121  =  0.  We 
find  by  inspection  and  trial  that  3  is  a  superior  limit  of  its  positive  roots. 
Hence  —  3  is  an  inferior  limit  of  the  negative  roots  of /(x)  =  0. 

820  On  rational  transformations  in  general.  If  we  eliminate  x 
between  the  equations  f(x)  =  0  and  y  =  —  x,  we  obtain 
y(—  y)=  0.  We  have  shown  in  §  811  tliat  the  roots  y  of 
/(—  y)=  0  are  connected  with  the  roots  x  off(x)  =  0  by  the 
relation  y  =  —  x.  This  is  an  illustration  of  the  general  theo- 
rem that  if  we  properly  eliminate  x  between  f{x)  =  0  and  any 
equation  of  the  form  y  =  ^(x),  where  (f>(x)  is  rational,  we 
shall  obtain  an  equation  F  (y)  =  0  whose  roots  are  connected 
with  those  of  f(x)  =  0  by  tlie  relation  y  =  <^  (x),  so  that  if  the 
roots  of /(a-)  =  0  are  ^i,  (3.,,  ■■■,  fi,,  those  of  F{y)  =  0  are  <^  (ySi), 
<^(A)^  •••,  <^(^„).  The  transformations  of  §§812,  814,  817 
afford  further  illustrations  of  this  theorem.  In  the  first  of 
these  transformations  the  equation  y  =  <f>  (x)  is  y  =  kx,  in  the 
second  it  is  ?/  =  1  /a-,  and  in  the  third  it  is  y  =  x  —  k.  When, 
as  in  these  cases,  y  =^  cf)(x)  can  be  solved  for  x,  the  elimination 
of  X  is  readily  effected. 

Example  1.  Find  the  equation  whose  roots  are  the  squares  of  the 
roots  of  x^  +  px"^  +  qx  +  r  =  0. 

In  this  case  the  relation  y  =  <l){x)  is  y  =  x^. 

Solving  y  =  x?  for  x,  we  have  x  =  ±  V^.  And  substituting  ±  y/y  for 
X  in  the  given  equation  and  rationalizing,  we  obtain 

'/  +  (2  9  -  p2)  y2  +  (^2  _  2pr)  y-r^  =  0. 

Example  2.  If  the  roots  of  x^  +  px"^  +  qx  +  r  =  0  are  cx.  /3,  y,  find  the 
equation  whose  roots  are  ^7,  ya,  aji. 

We  first  endeavor  to  express  each  of  the  proposed  roots  ^y,  ya, 
a^  in  terms  of  a  single  one  of  the  given  roots  a,  j8,  7  and  the  given 


THEORY    OF   EQUATIONS  443 

coefficients  p,  q,  r.  This  is  readily  done,  for,  since  —  r  =  a^y,  we  have 
^y  =  apy/a  =  -r/a,     ya  -  afiy / ^  =  -  r / ^,     a^  =  a^y /y  =- r / y. 

Hence  each  root  y  of  the  required  equation  is  connected  with  the  corre- 
sponding root  X  of  the  given  equation  by  the  relation  y  —  —  r /x. 

Solving  y  —  —  r/x  for  x,  we  have  x  =  —  r/y. 

And  substituting  —r/y  for  x  in  x^  +  px^  +  qx  +  r  =  0  and  simpli- 
fying, we  have  y^  —  qy'^  +  pry  —  r^  =  0,  which  is  the  equation  required. 

EXERCISE   LXXII 

1.  Change  the  signs  of  the  roots  of  x'  +  3  x*  -  2  x^  +  6  x  +  7  =  0. 

2.  Multiply  the  roots  of  2  x**  +  x^  -  4  x^  -  6  x  +  8  =  0  by  -  2.     Also 
divide  them  by  3. 

3.  In5xS  —  x^  +  Sx^  +  Ox+lO^O  replace  each  root  by  its  reciprocal. 

4.  Diminish  the  roots  of  2  x^  +  x*  -  3  x^  +  6  =  0  by  2.    Also  increase 
them  by  1. 

5.  Transform    the   equation  x*  -  x3/3  +  x2/4  +  x/25  -  1/48  =  0 
into  another  whose  coefficients  are  integers,  the  leading  one  being  1. 

6.  Transform  the  equation  3  x*  -  36  x'  +  x  -  7  =  0  into  another  which 
lacks  the  term  involving  x^. 

7.  Transform  the  following  into  equations  which  lack  the  x  term. 
(1)  x3  +  (i X,-  +  9x  +  10  =  0.  (2)  x3  -  x2  -  X  -  3  =  0. 

8.  If  the  roots  of  x*  +  x^  —  x  +  2  =  0  are  a,  /3,  7,  5,  find  the  equation 
whose  roots  are  a^,  /S^,  72,  5-. 

9.  If  the  roots  of  x*  +  3  x^  +  2  x^  -  1  =  0  are  o-,  /3,    y,  5,  find  the 
equation  whose  roots  are  |3  +  7  +  5,  a:  +  7  +  5,  a  +  i3  +  5,  a  +  jS  +  7. 

10.  If  the  roots  of  x^  +  px-  +  gx  +  r  =  0  are  a,  /3,  7,  find  the  equa- 
tions whose  roots  are 

^  '    7  '    cr  '     /3  '  ^"'  )3  +  7 '  7  +  a  '  a  +  /3 

11.  If  the  roots  of  x^  +  2x2  +  3x  +  4  =  0  are  a,  /3,  7,  find  the  equa- 
tions  whose  roots  are 

(1)  ^2  +  ^2^  ^2  +  a:2,  a:2  +  ^.         (2)  a{^  +  7),  ^  (7  +  «),  7  («  +  ^)- 

111  a  8  7 

(3)  ^7  +  -,  7«  +     .   ap  +  -•       (4)  ' ' 


444  A    COLLEGE    ALGEBRA 

12.  Find  superior  and  inferior  limits  of  the  real  roots  of  the  following 
equations. 

(1)  X*  +  3x3  -  13x2 -6x  + 28  =  0.     (2)  2x5  -  120x2  -  38x  +  27  =  q. 
(3)  X*  -  29x2  +  50x  +  12  =  0.  (4)  2x5  -  26  x^  +  60x2  _  92  =  0. 

(5)  X*  -  14 x3  +  44 x2  +  28x  -  92  =  0. 

(6)  3x«  -  35x3  +  77x2 -50x- 110^0. 

IMAGINARY   ROOTS.     DES CARTES 'S    RULE  OF   SIGNS 

821  Theorem.  Let  f  (x)  =  0  denote  an  equation  with  real  coeffi- 
cients. If  it  has  imaglnari/  roots,  these  occur  in  pairs;  that  is, 
■  if  &  +  ib  is  a  root,  a  —  ib  is  also  a  root. 

For  if  a  4-  ib  is  a  root  of  /(x)  =  0,  then  f{x)  is  divisible  by 
a;  _  (a  +  ib),  §  795 ;  and  we  shall  prove  that  a  —  ib  is  a  root 
if  we  can  show  that  f(x)  is  also  divisible  hy  x  —(a  —  ib),  or, 
what  comes  to  the  same  thing,  if  we  can  show  that  f{x)  is 
divisible  by  the  product  [.x  —  (a  +  ^7>)] [x  —(a  —  ib)]. 

This  product  has  real  coefficients,  for,  since  i^  =  —  1, 
lx-(a  +  ib)]  [x  -  (a  -  ib)]  =  (x-  a)2  +  b^ 

=  x'--2ax+  (a^  +  b^). 

Since  the  polynomials  f{x)  and  x^  —  2  ax  +  (a^  +  b'-)  have 
the  common  factor  x  —{a  +  ib),  they  have  a  highest  common 
factor.  This  highest  common  factor  must  be  either  x—{a  +  ib) 
ov  x^  —  2ax+  (a^  +  b^).  But  it  cannot  he  x  —  (a  +  ib),  since 
this  has  imaginary  coefficients,  whereas  the  highest  common 
factor  of  two  polynomials  with  real  coefficients  must  itself 
have  real  coefficients,  §  469.  Hence  the  highest  common  factor 
oif(x)  and  x^-2ax+  (a^  +  b^)  is  x^  -  2  ax  +  (a""  +  P)  ;  in 
other  words,  f(x)  is  divisible  hy  x^  -  2  ax  +  (a-  +  b-),  as  was 
to  be  demonstrated. 

Example.  One  root  of  2  x^  +  5  x^  +  4G  x  -  87  =  0  is  -  2  +  5  i.  Solve 
this  equation. 

Since  -  2  +  5  i  is  a  root,  —  2  —  5  i  is  also  a  root.  But  the  sum  of  all 
the  roots  is  —  5/2,  §  806.     Hence  the  third  root  is 

-6/2-(-2  +  5i-2 -5i)  =  3/2. 


THEORY   OF    EQUATIONS  445 

Corollary  1.     Every  polynomial  f  (x)  icith  real  coefficients  is     822 
the  product  of  real  factors  of  the  first  or  second  degree. 

For  to  each  real  root  c  of  /(a*)  =  0  there  corresponds  the 
real  factor  x  —  c  otf(x),  §  795 ;  and  to  each  pair  of  imaginary 
roots  a  +  ib,  a  —  ib  of  f(x)  =  0  there  corresponds  the  real 
factor  a:-  -  2  ax  +  (a^  +  b-)  oif(x),  §  821. 

Corollary  2.     The  product  of  those  factor's  o/f  (x)  which  corre-     823 
sjjond  to  the  imaginary  roots  of  f(x)  =  0  is  a  function  of  x 
%iihich  is  2JiJsitire  for  all  real  values  of  x. 

Tor  it  may  be  expressed  as  a  product  of  factors  of  the  form 
(x  —  ay  +  b'^,  §  821,  and  every  such  factor,  being  a  sum  of 
squares,  is  positive  for  all  real  values  of  x. 

Corollary  3.     Every  equation  with  real  coefficients  whose  degree     824 
is  odd  has  at  least  one  real  root. 

For  the  number  of  its  imaginary  roots,  if  it  have  any,  is 
even,  §  821,  and  the  total  number  of  its  roots,  real  and  imagi- 
nary, is  odd,  §  798.     Hence  at  least  one  root  must  be  real. 

Tims,  the  roots  of  a  cubic  equation  with  real  coefficients  are  either  all 
of  them  real,  or  one  real  and  two  imaginary. 

By  the  reasoning  employed  in  §  821  it  may  be  proved  that     825 
if  a  -\-  'wb  is  a  root  of  a  given  equation  with  rational  coeffi- 
cients, a  —  -y/b  is  also  a  root;  it  being  understood  that  a  and  b 
are  themselves  rational,  but  "vb  irrational. 

Irreducible  equations.     Let  (j>(x)  =  0  be  an  equation  whose     826 
coefficients   are  both   rational  and   real.     We  say  that   this 
equation  is  irreducible  if  <j>(x)  has  no  factor  whose  coefficients 
are  both  rational  and  real  (compare  §  486). 

Thus,  x2  —  2  =  0  and  x-  +  x  +  1  =  0  are  irreducible  equations,  but 
x~  —  4  =  0  is  not  irreducible. 

Theorem.     Let   t(x)=0  be   any  equation  ivhose   coefficients     827 
((re  both  rational  and  real,  and  let  ^(x)=  0  be  an  irreducible 
equation  of  the  same  or  a  lower  degree. 


446  A    COLLEGE   ALGEBRA 

If  one  of  the  roots  of  ^  (x)  =  0  he  a  root  of  f  (x)  =  0,  then  all 
of  the  roots  of  ^  (x)  =  0  are  roots  of  f  (x)  =  0. 

This  may  be  proved  by  the  reasoning  of  §  821.  For  if 
f(x)  =  0  and  <^  (x)  =  0  have  the  root  c  in  common,  f(x)  and 
<l>  (x)  have  the  common  factor  x  —  c,  ^  795,  and  therefore  a 
highest  common  factor  which  is  either  x  —  c,  some  factor  of 
<f>  (x)  which  contains  a;  —  c,  or  ^  (x)  itself. 

But  since,  by  hypothesis,  <f>(x)—  0  is  an  irreducible  equa- 
tion, <ji(x)  is  the  only  one  of  these  factors  Avhich  has  real  and 
rational  coefficients  such  as  the  highest  common  factor  oif(x) 
and  (f)  (x)  must  have,  §  469. 

Therefore  <^  (x)  is  itself  the  highest  common  factor  of  f(x) 
and  <t>(x)-  in  other  words, /(x)  is  exactly  divisible  by  ^(a;). 

Hence  f(x)  may  be  expressed  in  the  form  f(x)  =  Q<j)  (x), 
where  Q  is  integral,  and  from  this  identity  it  follows  that 
f{x)  vanishes  whenever  ^(r)  vanishes;  in  other  words,  that, 
every  root  of  <^  (a;)  =  0  is  a  root  of  f{x)  =  0. 

828  Permanences  and  variations.  In  any  polynomial  f(x),  or  equa- 
tion/(.r)  =  0,  with  real  coefficients  a,  jJermanence  or  continuation 
of  sign  is  said  to  occur  wherever  a  term  follows  one  of  like 
sign,  and  a  variation  or  change  of  sign  wherever  a  term  fol- 
lows one  of  contrary  sign. 

Thus,  in  x6^-  x*  -x3  +  2x2  +  3a;_l  =  o  permanences  occur  at  the 
terms  -  x^  and  3x,  and  variations  at  the  terms  -  x^,  2x2,  and  -  1. 

829  Theorem.  If  f  (x)  is  exactly  divisible  bij  x  —  h,  where  b  is 
positive  and  the  coeffi.cients  ofi(x)  are  real,  the  quotient  <f>(x) 
will  have  at  least  one  less  variation  than  f  (x)  has. 

For  since  b  is  positive,  it  follows  from  the  rule  of  synthetic 
division,  §  411,  that  when /(a:)  is  divided  by  a;  -  ^.,  the  coeffi- 
cients of  the  quotient  are  positive  until  the  first  negative 
coefficient  of  f(x)  is  reached.  If  then  or  later  one  of  them 
becomes  negative  or  zero,  they  continue  negative  until  the 
next  positive  coefficient  off(x)  is  reached,  and  so  on.  Hence 
<f>  (x)  can  have  no  variations  except  such  as  occur  at  the  same 


THEORY    OF    P:QUATI0NS  447 

or  earlier  terms  of /(a-).  But  since  the  division  is,  by  hypoth- 
esis, exact,  the  last  sign  in  <^  (x)  must  be  contrary  to  the  last 
sign  in  f{x),  and  therefore  ^  (x)  must  lack  the  last  variation 
in/(,r). 

1     +1     -2     -10     -1     +12     -4[2  Thus,  /(x)  =  a;6  +  a;5  -  2  x* 

2     -f6     +    8     -  4     -  10     +  4  -  10  x^  -  x2  +  12  X  -  4  is  ex- 

1+3+4-2-5+2,  0  actly  divisible  by  x  -  2,  tlie 
quotient  being  0  (x)  =  x^  +  3  x*  +  4  x^  —  2  x^  —  5  x  +  2.  Observe  that  the 
first  two  variations  of/  (x)  ai-e  reproduced  in  </>  (x),  but  not  the  third. 

1      -1     +1     -7     +2|^  Again,  f(x)  =  x*  -  x-''  +  x'^  -  7  x  +  2   is 

2     +2_     +_n     —2_         exactly   divisible    by   x  —  2,    the   quotient 

1     +1     +3     —1,        0  being  0(x)  =  x^ +  x2  + 3x —  1.     In  this  case 

only  one  of  the  four  variations  of /(x)  is  reproduced  in  0(x),  and  we 
have  an  illustration  of  the  fact  that  when  intermediate  variations  of  f{x) 
disappear  in  0(x),  they  disappear  in  pairs. 

Theorem  (Descartes's  rule  of  signs).   A71  eq uation  f  (x)  =  0  cannot     830 
have  a.  greater  number  of  positive  roots  than  it  has  variations,  7ior 
a  greater  number  of  negative  roots  than  the  equation  f  (—  x)  =  0 
has  variations. 

4].  For  let  /3i,  yS.j,  •■  -,  (3,.  denote  the  positive  roots  of /(.r)  =  0. 
if  we  divide  /(if)  by  x  —  /3i,  the  quotient  thus  obtained 
by  X  —  (32,  and  so  on,  we  obtain  a  final  quotient  c{>(x)  which 
has  at  least  r  less  variations  than  /(.r)  has,  §  829.  Therefore, 
since  </>  (x)  cannot  have  less  than  no  variations,  f(x)  must  have 
at  least  r  variations,  that  is,  at  least  as  many  as  f(x)  =  0  has 
positive  roots; 

2.  The  negative  roots  of/(.r)=0  become  the  positive  roots 
of  /(—  x)  =  0,  §  811.  And,  as  just  demonstrated,  /(—  x)  =  0 
cannot  have  more  positive  roots  than  variations.  Hence 
/(a-)  =  0  cannot  have  more  negative  roots  than  /(— a-)=0 
has  variations. 

Thus,  the  equation  /(x)  =  x^  —  x^  —  x'  +  x  —  1  =  0  cannot  have  more 
than  three  positive  roots  nor  more  than  one  negative  root.  For  /(x)  =  0 
has  three  variations,  and  /(-  x)  =  0,  that  is,  x^  +  x^  +  x^  -  x  -  1  =  0,  has 
one  variation. 


448  A   COLLEGE   ALGEBRA 

831  Corollary.     A  complete  equation  cannot  have  a  greater  nuin^ 

her  of  negative  roots  titan  it  has  j)ernia?ie?ices. 

For  when  f(x)  =  0  is  complete  its  permanences  correspond 
one  for  one  to  the  variations  of /(—  x)  =  0,  since  of  every  two 
consecutive  like  signs  in  /(x)  =  0  one  is  changed  in  /(—  x)  —  0, 
§811. 


Thus, 

if     f{x)  =  0  is  x5  +  X*  -  6  x3  -  8  X'-  -  7  X  +  1  =  0, 

(1) 

en 

/( -  x)  =  0  is  x5  -  X*  -  6  x^  +  8  X-  -  7  X  -  1  =  0. 

(2) 

In  (1)  we  have  permanences  at  the  terms  x*,  —  Sx^,  —  7x,  and  at  the 
corresponding  terms  of  (2),  namely,  —  x'*,  8  x'-,  —  7  x,|  we  have  variations. 

Since  (1)  has  two  variations  and  three  permanences,  /(x)  =  0  cannot 
have  more  than  two  positive  roots  nor  more  than  three  negative  roots. 

832  Detection  of  imaginary  roots.  In  the  case  of  an  incomplete 
equation  we  can  frequently  prove  the  existence  of  imaginary 
roots  by  aid  of  Descartes's  rule  of  signs. 

Let  f  (x)  =  0  be  an  equation  of  the  nth  degree  which,  has  no 
zero  roots,  and  let  v  and  v'  denote  the  number  of  variations  in 
f  (x)  =  0  and  f  (—  x)  =  0  respectively.  The  equation  f  (x)  =  0 
must  have  at  least  n  —  (v  +  v')  imaginary  roots. 

For  f{x)  —  0  cannot  have  more  than  v  positive  roots  nor 
more  than  v'  negative  roots,  §  830,  and  therefore  not  more 
than  w  +  v'  real  roots  all  told.  The  rest  of  its  n  roots  must 
therefore  be  imaginary. 

This  theorem  gives  no  information  as  to  the  imaginary  roots 
of  a  complete  equation,  since  v  +  v'  is  equal  to  n  in  such  an 
equation. 

Example.     Show  that  x^  +  x^  +  1  =  0  lias  four  imaginary  roots. 

In  this  case /(x)  =  0  isx^  +  x2  +  1  =  0,  and/(-x)  =  0  is  x5-x2-l  =  0. 

Hence  n  —  (u  +  u')  =  5  —  (0  +  1)  =  4,  so  that  there  cannot  be  less  than 
four  imaginary  roots.  But  since  there  are  five  roots  all  told  and  one  of 
them  is  real,  the  degree  of  x^  +  ic^  +  1  =  0  being  odd,  §  824,  there  cannot 
be  more  than  four  imaginary  roots.  Hence  x»  +  x'-  +  1  =  0  has  exactly 
four  imaginary  roots. 


THEORY    OF    EQUATIONS  449 

EXERCISE   LXXIII 

1.  One  root  of  2 x*  -  x-'  +  5x-  +  13 x  +  5  =  0  is  1  -2i.  Solve  this 
equation. 

2.  One  root  of  2  x*  -  11  x^  +  17  x-  -  10  x  +  2  =  0  is  2  +  V2.  Solve 
this  equatinii. 

3.  Find  the  equation  of  lowest  degree  with  rational  coefficients  twc 
of  whose  roots  are  —  5  +  2  i  and  —  1  +  v  5. 

4.  Find  tlie  irreducible  equation  one  of  whose  roots  is  V2  +  i. 

5.  What  conclusions  regarding  the  roots  of  the  following  equations 
can  be  drawn  by  aid  of  Descartes's  rule  and  §  8o2? 

(1)  X*  +  1  =  0.  (2)  X*  -  x2  -  1  =  0. 

(3)  x<  +  2  x^  +  x'-;  +  X  +  1  =  0.  (4)  X*  -  2  x'  +  x^  -  X  +  1  =  0. 

(5)  X'  +  xs  +  x^  -  X  +  1  =  0.  (6)  x"  +  x^  -  x2  -1=0. 

(7)  x5  -  4  x--^  +  3  =  0.  (8)  x""  -  x2»  +  X"  +  x  +  1  =  0. 

6.  Show  that  a  complete  equation  all  of  whose  roots  are  real  has 
as  many  positive  roots  as  variations,  and  as  many  negative  roots  as 
permanences. 

7.  Given  that  all  the  roots  of  x^  +  3  x*  -  1 5  x^  -  35  x^  +  54  x  +  72  =  0 
are  real,  state  how  many  are  positive  and  how  many  negative. 

8.  Prove  by  Descartes's  rule  that  x^"  +  1  =  0  has  no  real  root.  What 
conclusions  can  be  drawn  by  aid  of  this  rule  regarding  the  roots  of 
x2«+i  +  l=0?   x2«-l  =  0?   x2''  +  i-l  =  0? 

9.  Prove  that  an  equation  which  involves  only  even  powers  of  x  with 
positive  coefficients  cannot  have  a  positive  or  a  negative  root. 

10.  Prove  that  an  equation  which  involves  only  odd  powers  of  x  with 
positive  coefficients  has  no  real  root  except  0. 

11.  Show  that  the  equation  x^  +  px  +  q  =  0,  where  p  and  q  are  posi- 
tive, has  but  one  real  root,  that  root  being  negative. 

12.  Show  that  an  incomplete  equation  which  has  no  zero  roots  must 
have  two  or  more  imaginary  roots  except  when,  as  in  x*  —  3  x^  +  1  =  0,  the 
missing  terms  occur  singly  and  between  terms  which  have  contrary  signs. 

13.  Show  that  in  any  equation /(x)  =  0  with  real  coefficients  there 
must  be  an  odd  number  of  variations  between  two  non-consecutive 
contrary  signs,  and  an  even  number  of  variations,  or  none,  between  two 
non-consecutive  like  signs. 


450  A    COLLEGE    ALGEBRA 

14.  Prove  that  in  the  product  of  the  factors  corresponding  to  the 
negative  and  imaginary  roots  of  an  equation  with  real  coefficients  the 
final  term  is  always  positive,  and  then  show  that  if  this  product  has 
any  variations  their  number  is  even. 

15.  Prove  that  when  the  number  of  variations  exceeds  the  number  of 
positive  roots,  the  excess  is  an  even  number. 

16.  Show  that  x*  +  x^  —  x^  +  x  —  1  =  0  has  either  one  or  three  posi- 
tive roots  and  one  negative  root. 

17.  Show  that  every  equation  of  even  degree  whose  absolute  term  is 
negative  has  at  least  one  positive  and  one  negative  root. 

LOCATION   OF   IRRATIONAL   ROOTS 

833         Theorem  1.     //'f(a)  atul  f(b)  liavc  contrari/  signs,  a  root  of 
f  (^xj=  0  lies  between  a  and  b. 

This  may  be  proved  as  in  the  following  example.  A  general 
statement  of  the  proof  will  be  given  subsequently. 

Example.  Prove  that  /(x)  =  x^  —  3  x  +  1  =  0  has  a  root  between  1 
and  2. 

The  sign  of /(I)  =  —  1  is  minus  and  that  of /(2)  =  .3  is  plus. 

By  computing  the  values  of /(x)  for  x  =  1.1,  1.2,  1.3,  •  •  •  successively, 
we  find  two  cmsecutive  tenths  between  I  and  2,  namely,  1.5  and  1.6,  for 
which  /(x)  has  the  same  signs  as  for  x  =  1  and  x  =  2  respectively ;  for 
/(1. 5)  =-  .125  is  minus,  and  /(l.O)  =  .20(i  is  plus. 

By  the  same  meth(5d  we  find  two  consecutive  hundredths  between  1.5  and 
1.0,  namely,  1.53  and  1.54,  for  which/(x)  has  the  same  signs  as  for  x  =  1  and 
X  =  2  ;  for /(I. .53)  =  -  .008423  is  minus  aml/(1.54)  =  .032204  is  plus. 

This  proce.ss  may  be  continued  indefiniiely.  It  determines  the  two 
never-ending  sequences  of  numbers: 

(a)  1,  1.5,  1.53,  1.532,  •  •  •  (b)  2,  1.0,  1.54,  1.533,  •  • ., 

the  terms  of  which  approach  the  .same  limiting  value,  §§  192,  193.     Call 
this  limiting  value  c.     It  is  a  root  of /(x)  =  0,  that  is,  f{c)  =  0. 

For,  by  §  509,  if  x  be  made  to  run  through  either  of  the  sequences  of 
values  (a)  or  (b),  f(x)  will  approach  f(c)  as  limit.  But  since/(x)  is  always 
negative  as  x  runs  tlirough  the  sequence  (a),  its  limit /(c)  cannot  be  posi- 
tive ;  and  since /(x)  is  always  positive  as  x  runs  through  the  sequence  (b), 
its  limit /(c)  cannot  be  negative.     Hence /(c)  is  zero. 


THEORY    OF    EQUATIONS  451 

Theorem  2.     If  neAther  a  nor  b  is  a  root  of  f  (x)=  0,  and  an     834 
odd  number  of  the  roots  of  i(x)  =  0  lie  between  a  and  b,  f  (a)  and 
f(b)  have  contrary  signs;  but  if  no  root  or  an  even  number  of 
roots  lie  between  a  and  b,  f  (a)  and  f  (b)  ha^te  the  same  sign. 

Conversely,  if  f(a)  and  f  (b)  have  contrary  signs,  an  odd 
nu77iber  of  the  roots  of  f  (x)  =  0  lie  between  a  and  b  ;  but  (/"  f  (a) 
and  f  (b)  have  the  same  sign,  either  no  root  or  an  even  number 
of  roots  lie  between  a  and  b. 

Suppose  that  a  <b  and  that  y3i,  yS.,,  •  •  •,  /S^  is  a  complete  list 
of  the  roots  of /(a-)  =  0  between  a  and  b.  Then/(x)  is  exactly 
divisible  by  (x  -  y8i)  {x  -  y8,)  ■  ■  ■  {x  -  ft,.),  §  418,  and  if  we  call 
the  quotient  ^  (x),  we  have 

/(,r)  =  (:«  -  A)  (.^  -  i3,)  ■  ■  •  (.r  -  /?,.)  <^  (x).  (1) 

Substituting  first  a  and  then  b  for  a-  in  (1)  and  dividing  the 
first  result  by  the  second,  we  obtain 

f(a)       a-  fti    a-  ft.       a  -  ft,.   fj>(a) 


f{l>)       b-ft,    b-ft,       b~ft,.    4>{b)  (^^ 

In  the  product  (2)  the  factor  ^  (a. )  /  c^  (/v)  is  positive.  For 
^  (a)  and  <^  (b)  have  the  same  sign,  since  otherwise,  by  §  833, 
between  a  and  b  there  would  be  a  root  of  ^  (x)  =  0  and  there- 
fore, by   (1),  a  root  of  f{x)  =  0  in  addition  to  the  roots  ft^, 

On  the  other  hand,  each  of  the  r  factors  {a  —  fti)/(b  —  fti), 
and  so  on,  is  negative,  since  each  of  the  r  roots  fti,  ft.2,  ■  ■  ■,  /S^  is 
greater  than  a  and  less  than  b. 

Therefore,  when  r  is  odd, /(rt)//(^)  is  negative,  that  is, 
f(a)  and  f(b)  have  contrary  signs  ;  but  when  r  is  even  or 
zero,  f{(i)/f{b)  is  positive,  that  is,  /(«)  and  f(l))  have  the 
same  sign. 

Conversely,  when  f(a)  and  f(b)  have  contrary  signs,  so  that 
f(c-)/f{b)  is  negative,  it  follows  from  (2)  that  r  is  odd;  and 
when  f(a)  and  f(b)  have  the  same  sign,  it  follows  that  r  is 
even  or  zero. 


452  A   COLLEGE   ALGEBRA 

835  Observe  that  in  the  proofs  of  the  preceding  theorems,  §§  833,. 
834,  no  use  has  been  made  of  the  assumption  that  every  equa- 
tion/(x)=  0  has  a  root.  Kotice  also  that  in  applying  these 
theorems  a  multiple  root  of  order  r  is  to  be  counted  as  r 
simple  roots. 

From  §  834  it  follows  that  as  x  varies  from  a  to  h,  f(x)  will 
change  its  sign  as  x  passes  through  each  simple  root  or  multiple 
root  of  odd  order  oi  f(x)  =  0  which  lies  between  a  and  b,  and 
that /(a;)  will  experience  no  other  changes  of  sign  than  these. 

Thus,  iif{x)  =  (x  -  2)  (x  -  3)'-(x  -  4)^,  and  x  be  made  to  vary  from  1 
to  5,  the  sign  of /(x)  will  be  plus  between  x  =  1  and  x  =  2,  minus  between 
X  =  2  and  x  =  4,  and  plus  between  x  =  4  and  x  =  5. 

836  Location  of  irrational  roots.  By  aid  of  the  theorem  of  §  833 
it  is  usually  possible  to  determine  between  what  pair  of  con- 
secutive integers  each  of  the  fractional  and  irrational  roots  of 
a  given  numerical  equation  lies. 

Example.     Locate  the  roots  of  /(x)  =  x*  -  6  x^  +  x2  +  12  x  -  6  =  0. 

By  Descartes's  rule  of  signs,  §  830,  this  equation  cannot  have  more 
than  three  positive  roots  nor  more  tlian  one  negative  root. 

To  locate  the  positive  roots  we  compute  successively /(O),  /(I),  /(2),  • . . 
until  three  roots  are  accounted  for  by  §  83.3  or  until  we  reach  a  value  of 
X  which  is  a  superior  limit  of  the  roots,  §  803. 

Thus,  using  the  method  of  synthetic  division,  as  in  §  414,  we  find 
/(O)  =  -  6,  /(I)  =  2,  /(2)  =  -  10,  /(.3)  =  -  42,  /(4)  =  -  70,  /{5)  =  -  46, 
/(O)  =  102. 

Hence,  §  833,  one  root  lies  between  0  and  1,  another  between  1  and  2, 
and  the  third  between  5  and  6.  There  cannot  be  more  than  one  root  in 
any  of  these  intervals,  since  there  are  only  three  positive  roots  all  told. 

Making  a  similar  search  for  the  negative  root,  we  have/{0)=-6, 
/(-!)= -10, /(-2)  =  38.    Hence  the  negative  root  lies  between  -land -2. 

The  mere  substitution  of  integers  for  x  in  f(x)  will  of  course 
not  lead  to  the  detection  of  all  the  real  roots  when  two  or  more 
of  them  lie  between  a  pair  of  consecutive  integers.  This  case 
will  be  considered  in  §  844  and  again  in  §  8G4,  where  a  method 
is  given  for  determining  exactly  how  many  roots  lie  between 
any  given  pair  of  numbers. 


THEORY    OF   EQUATIONS  458 

EXERCISE   LXXIV 

Locate  the  real  roots  of  each  of  the  following  equations. 
1.  2  x3  -  3  x2  -  9  X  +  8  =  0.  2.    x«  +  x2  -  4  x  -  2  =  0. 

3.    x3  -  3  x^  -  2  X  +  5  =  0.  4.  2  x3  +  3  x2  -  10  X  -  15  =  0. 

5.    x3  -  4x2 -4x  + 12  =  0.  6.    x*  +  13x2  +  54x  +  71  =  0. 

7.   x5  +  5  X  +  19  =  0.  8.  X*  -  95  =  0. 

9.   X*- 8x3 +  14x2 +  4x -8  =  0.     IQ.  x*  +  Sx^  +  x2  -  13x  - 7  =  0. 

11.  X*  -  11  x5  +  32  x2  -  4  x  -  46  =  0. 

12.  x5  +  2x*  -  16x3  _  24x2  +  48x  +  32  =  0. 

13.  Assuming  that  when  x  is  very  large  numerically  the  sign  of  /(x) 
is  that  of  its  term  of  highest  degree,  show  that 

(1)  Every  equation  x"  +  6iX"-i  +  . . .  +  6„  =  0  with  real  coefficients, 
in  which  n  is  even  and  b,,  is  negative,  has  at  least  one  positive  and  one 
negative  root. 

(2)  The  four  roots  of  the  equation 
A;2  (x  —  6)  (x  —  c)  +  I-  (x  —  c)  (x  —  a) 

+  7/(2  (^  _  (j)  ^x  _  5)  _  X  (x  —  a)  (x  —  6)  (x  —  c)  =  0 
lie  between  —  co  and  a,  a  and  &,  b  and  c,  c  and  cc  respectively,  it  being 
assumed  that  a,  6,  c,  k,  I,  m  are  real  and  that  a<b<c. 

14.  Show  that  every  equation  of  the  form  x^  +  (x  —  1)  (ax  —  1)  =  0, 
where  a>3,  has  two  roots  between  0  and  1,  namely,  one  between  \/a 
and  1  —  1/a  and  one  between  1  —  1/a  and  1. 

15.  Show  that  x*  +  (x  —  1)  (2  x  —  1)  (ax  —  1)  =  0,  where  a  >  5,  has  roots 
between  0  and  1/a,  1/a  and  1  —  2/a,  1  —  2/a  and  1. 

COMPUTATION   OF   IRRATIONAL   ROOTS 

Horner's  method.  Positive  roots.  There  are  several  methods  837 
by  which  approximate  values  of  the  irrational  roots  of  numer- 
ical equations  can  be  computed.  The  most  expeditious  of 
these  methods  is  due,  in  its  perfected  form,  to  an  English 
mathematician  named  Horner.  It  may  best  be  explained  in 
connection  with  an  example. 


15 

-59L3 

21 
6 

18 
-41 

39 

45 

454  A    COLLEGE   ALGEBRA 

Example.     Find  the  positive  root  of /(x)  =  2x3  _|-  x2  —  I5x  —  59  =  0, 
1.    By  the  method  of  §  836,  we  find  that  the  required  root  lies  between 
3  and  4.     Hence  if  it  be  expressed  as  a  decimal  number,  it  will  have  the 
form  3.  ^yS  •  •  • ,  where  jS,  y,  S,  ■  ■  ■  denote  its  decimal  figures. 
2     +    1     -  15     -  59  [3         2.    Diminish  the  roots  of  f{x)  =.  0  (1)  by  3. 
We  obtain  the  transformed  equation 

<^(x)  =  2x3  + 19x^  + 45x  -  41  =  0,      (2) 
which  has  the  root  .^y5  ■  ■  ■  lying  between  0 
13         45  and  1. 

5  Testing  x  =  .1,  .2,  .3,  ••  •  in  4>(x),  we  find 

19  that  0(.6)  is  -  and  0(.7)  is  +.     Hence  the 

root  of  (2)  lies  between  .6  and  .7,  that  is,  ^3  is  6,  and  the  root  of  (1) 
to  the  first  decimal  figure  is  3.6. 

2+19        +45  -  41  1^  3.    Diminish  the  roots  of  (2)  by  .6. 

We  obtain 
i/'(x)  =  2x5  +  22.6x5 

+  69.96X- 6.728  =  0,  (3) 
which  has  the  root  .0  ^5  •  ■  •  lying  between 
0  and  .1. 

Te.sting  x  =  .01,  .02,  ■  ■  ■  m  ^p  (x),  we 
find  that  \p  (.09)  is  —  and  i/-  (.1)  is  + .    Hence  the  root  lies  between  .09  and 
.1,  that  is,  7  is  9,  and  the  root  of  (1)  to  the  second  decimal  figure  is  3.69. 
4.    Diminish  the  roots  of  (3)  by  .09,  and  so  on. 

838  The  reckoning  may  be  conducted  more  simply  than  in  this 
example,  as  will  be  shown  in  the  following  sections.  But 
before  turning  from  the  example,  observe  that  the  absolute 
terms  of  the  first  and  second  transformed  equations,  (2)  and 
(3),  namely,  —  41  and  —  6.728,  have  the  same  sign.  This  is  as 
it  should  be,  since  -  41  =  <^  (0)  and  -  6.728  =  ^  (.6).  For  were 
^(0)  and  ^(.6)  to  have  contrary  signs,  the  root  of  <ji(x)=  0 
would  lie  between  0  and  .6,  §  833,  and  therefore  .6  would  not  be 
its  first  figure.  The  like  is  true  of  the  subsequent  transformed 
equations,  and,  in  general, 

When  the  given  equation  has  but  one  root  with  the  integral 
part  a,  the  absolute  terms  of  all  the  transformed  equations  used 
in  finding  this  root  will  have  the  same  sign,  if  the  reckoning  is 
correctly  performed. 


+  19 

+  45 

-41|.6 

1.2 

12.12 

34.272 

20.2 

57.12 

-    6.728 

1.2 

12.84 

21.4 

69.96 

1.2 

22.6 

THEORY   OF    EQUATIONS  455 

In  the  example  we  found  the  first  figure  of  the  root  of  each  839 
transformed  equation,  that  is,  the  successive  decimal  figures  of 
the  root  of  the  given  equation,  by  the  method  of  substitution. 
But  the  first  figure  of  the  root  of  each  transformed  equation 
from  the  second  on  may  ordinarily  be  found  by  merely  dividing 
the  absolute  term  of  the  equation,  with  its  sign  changed,  by  the 
coefficient  of  x.     This  is  called  the  viethod  of  trial  divisor. 

Thus,  consider  the  second  transformed  equation  in  the  example 

V'(x)  =  2x3  +  22.6 X-  +  69.90 X  -  6.728  =  0.  (3) 

This  equation  is  known  to  have  a  root  c  wliich  is  less  than  .1.  The 
second  and  higher  powers  of  such  a  number  c  will  be  much  smaller  than 
c  itself.  Thus,  even  (.09)2  jg  i^yf  .0081.  Hence,  were  c  known  and  sub- 
stituted in  (3),  the  first  two  terms  of  the  resulting  numerical  identity 

2  c3  +  22.6  c2  +  69.96  c  -  6.728  =  0 
would  be  very  small  numbers  in  comparison  with  the  last  two. 

Therefore  c  is  not  likely  to  differ  in  its  first  figure  from  the  root  of  the 
equation  69.96  x  -  6.728  =  0  (3') 

obtained  by  discarding  the  x^  and  x^  terms  in  (3). 

But  solving  (3'),  we  have  x  =  6.728/69.96  =  .09  +,  that  is,  we  find,  as 
above,  that  the  first  figure  of  the  root  of  (3)  is  9. 

This  method  cannot  be  trusted  to  give  the  first  figure  of  the 
root  of  the  first  transformed  equation  correctly.  But  it  will 
usually  give  at  least  some  indication  as  to  what  that  figure  is 
and  so  lessen  the  number  of  tests  that  need  to  be  made  in 
applying  the  method  of  substitutions.  Occasionally  the  method 
fails  to  give  correctly  the  first  figure  of  the  root  of  even  the 
second  transformed  equation.  But  in  such  a  case  the  error  is 
readily  detected  in  carrying  out  the  next  transformation ;  for 
if  the  figure  is  too  large,  a  change  of  sign  will  occur  in  the 
absolute  term  of  this  next  transformed  equation,  §  838  ;  if  it  is 
too  small,  the  first  figure  of  the  root  of  this  equation  will  be  of 
too  high  a  denomination. 

We  may  avoid  the  troublesome  decimals  which  occur  in  the     840 
transformations  after  the  first  by  multiplying  the  roots  of  each 
transformed  equation  by  ten,  §  812,  before  making  the  next 


456  A  collegp:  algebra 

transformation.  This  may  be  done  by  affixing  one  zero  to  the 
second  coefficient  of  the  equation  in  question,  two  zeros  to  its 
third  coefficient,  and  so  on.  We  then  treat  the  figure  of  the  root 
employed  in  the  next  transformation  as  if  it  were  an  integer. 

Thus,  the  first  transformed  equation  in  the  example  in  §  837  was 

2x3  +  19x2 +  45x- 41  =  0,  (2) 

and  we  found  that  it  had  a  root  of  the  form  .6  +. 
Multiplying  the  roots  of  (2)  by  10,  we  obtain 

2  x3  +  190  x2  +  4500  X  -  41000  =  0,  (20 

which  has  a  root  of  the  form  6  + . 

Diminishing  the  roots  of  (2')  by  6,  the  reckoning  differing  from  that 
above  given  only  in  the  absence  of  decimal  points,  we  have 

2  x3  +  226  x2  +  G990  x  -  6728  =  0,  (30 

whose  roots  are  ten  times  as  great  as  those  of 

2  x3  +  22.6 x2  +  69.96  X- 6. 728  =  0.  (3) 

The  method  of  trial  divisor  gives  .9  +  as  the  root  of  (3')  and  therefore, 
as  above,  .09  +  as  the  root  of  (3). 

841  We  may  now  arrange  the  reckoning  involved  in  computing  the 

root  of  2  x^  +  cc^  —  15  a;  —  59  =  0  to  the  thii'd  decimal  figure  as 
follows  : 

2+1         -  15  -  59  13.693 

6  21  18 

7  6  -41 

6  39 

13  45 


2  +190 

+  4500 

-  41000  [6 

12 

202 

1212 
5712 

34272 

-  6728 

12 
214 

1284 
6996 

6728 
6996 

.9  + 

12 

2  +  2260 

+  699600 

-  6728000  [9 

18 

20502 

6480918 

2278 

720102 

-   247082 

18 

20664 

2296 

740766 

18 
2314 

247082 
740766 

.3  + 

THEORY    OF    EQUATIONS  457 

Observe  tliat  here  each  figure  obtained  by  the  trial  divisor  method  is  a 
tenth,  thus,  .9,  .3.  Had  the  last  coefficient  of  the  second  transformed  equa- 
tion been  -  672  instead  of  -  6728,  we  should  have  had  672/6996  =  .09 
for  the  next  two  figures.  The  root  as  far  as  computed  would  then  have 
been  3.609  instead  of  3.69,  and  before  performing  the  next  transformation 
we  should  have  multiplied  the  roots  of  this  second  transformed  equation 
by  100,  that  is,  we  should  have  affixed  two  zeros  to  its  second  coefficient, 
four  to  the  third,  and  six  to  the  fourth. 

This  process  may  be  continued  indefinitely.  But  we  soon 
encounter  very  large  numbers,  and  after  a  few  decimal  figures 
of  the  root  have  been  obtained  we  can  find  as  many  more  as 
are  likely  to  be  required,  with  much  less  reckoning,  by  the 
following  contracted  method. 

The  last  transformed  equation  in  the  reckoning  above  given  is 

2  x^  +  2314  x2  +  740766  x  -  247082  =  0.  (4) 

Instead  of  affixing  zeros  to  the  coefficients  in  order  to  multiply  the  roots 
of  (4)  by  10,  we  may  substitute  x/10  for  x  in  (4),  §  812,  thus  obtaining 
.002  x^  +  23. 14  x'^  +  74076.6  x  -  247082  =  0.  (4') 

Ignoring  the  decimal  parts  thus  cut  off  from  the  coefficients  as  being 
too  small  to  affect  the  next  few  figures  of  the  root,  but  adding  1  to  the 
corresponding  integral  part  when  the  decimal  part  is  .5  or  greater,  we 
have  the  quadratic       23  x^  +  74077  x  -  247082  =  0.  (4") 

We  may  then  continue  the  reckoning  as  follows : 
23     +74077     -247082  1.003332 


69 

222438 

74146 
69 

-  24644 

?3 

+  7422^ 

-  24644 
22266 

742^ 

-   2378 
2226 

74^ 

152 
148 

That  is,  we  diminish  the  roots  of  the  quadratic  (4")  by  3  and  thus 
obtain  the  transformed  equation  23x2  _|.  74215x  -  24644  =  0  (5).  By  the 
method  of  trial  divisor,  we  find  that  the  next  figure  of  the  root  is  also  3. 


458  A    COLLEGE    ALGEBRA 

Before  performing  the  next  transformation  we  cut  off  figures  as  before 
and  thus  reduce  (5)  to  the  simple  equation  7422  x  —  24644  =  0.  The 
next  two  figures  of  the  root,  namely,  3,  2,  are  tlien  obtained  by  merely 
dividing  24644  by  7422  by  a  contracted  process  which  consists  in  cutting 
off  figures  at  the  end  of  the  divisor  instead  of  affixing  zeros  at  the  end  of 
the  dividend. 

843  Negative  roots.  To  find  a  negative  irrational  root  of /(x)  =  0, 
find  the  corresponding  positive  root  of  /(—  x)  =  0  and  then 
change  its  sign. 

Example.     Find  the  negative  root  of /(x)  =  x^  +  x^  -  lOx  +  9  =  0. 

Here/(—  x)  =  0  is  x^  —  x'^  —  lOx  —  9  =  0.  Its  positive  root,  found  by 
Horner's  method,  is  4.03293  approximately.  Hence  the  negative  root 
of  /(x)  =  0  is  —  4.03293  approximately. 

844  Roots  nearly  equaL  If  the  given  equation  has  two  roots 
lying  between  a  pair  of  consecutive  integers,  they  may  be 
found  as  in  the  following  example. 

Example.     Find  the  positive  roots,  if  any,  of /(x)  =  x3  +  x2— 10x  +  9  =  0. 

We  find  that/(0)  =  9,  /(I)  =  1,  /(2)  =  1,  /(3)  =  15,  and  the  reckoning 
shows  that  3  is  a  superior  limit  of  the  roots,  §  803.  Hence,  §  834,  either 
there  is  no  positive  root,  or  there  are  two  such  roots  both  lying  between 

0  and  1,  or  between  1  and  2,  or  between  2  and  3.  But  /(I)  and  /(2) 
differ  less  from  0  than  /(O)  and  /(3)  do.  Hence,  if  two  roots  exist,  we 
may  expect  to  find  them  between  1  and  2  rather  tlian  between  0  and  1  or 
between  2  and  3. 

We  therefore  diminish  the  roots  of /(x)  =  0  by  1,  obtaining 
0  (X)  =  x3  +  4  x2  -  5  X  +  1  =  0, 

which  has  two  roots  between  0  and  1  if  /(x)  =  0  has  two  roots  between 

1  and  2. 

Computing  the  values  of  0  (x)  for  x  =  .1,  .2,  .3,  •  •  • ,  we  find  that  0  (.2)  is 
+  and  (/,(.3)  is  -,  also  that  ^^(.7)  is  -  and  <!>{.%)  is  +.  Hence  0{x)  =  0 
has  a  root  between  .2  and  .3  and  another  between  .7  and  .8.  By  Horner's 
method  we  find  that  tliese  roots  are  .25560  and  .77733  approximately. 

Hence /(x)  =  0  has  the  two  positive  roots  1.2556  and  1.77733. 

845  On  locating  large  roots.  Tn  case  the  given  equation  f(x)  =  0 
has  a  root  which  is  greater  than  ten,  we  may  employ  the  follow- 
ing method  for  finding  the  figures  of  its  integral  part. 


THEORY    OF    EQUATIONS  459 

To  obtain  the  first  figure,  compute  the  values  of  f{x)  for 
X  =  10,  20,  •  •  -,  or,  if  necessary,  for  x  =  100,  200,  •  •  •,  and  so 
on,  applying  §  833.  Thus,  if  we  found  that/(400)  and  /(500) 
had  contrary  signs,  so  that  the  root  lay  between  400  and  500, 
the  first  figure  would  be  4.  To  find  the  remaining  figures, 
make  successive  transformations  of  the  equation,  as  when 
finding  the  decimal  figures.  Thus,  in  the  case  just  cited  we 
should  diminish  the  roots  of /(cc)  =  0  by  400  and  so  obtain  an 
equation  </>(a;)  =  0  having  a  root  between  0  and  100.  If  we 
found  that  this  root  lay  between  70  and  80,  the  second  figure 
of  the  root  v/ould  be  7.  We  should  then  diminish  the  roots  of 
(f)(^x)=  0  by  70  and  so  obtain  an  equation  ij/(x)=0  having  a 
root  between  0  and  10.  If  we  found  that  this  root  lay  between 
8  and  9,  we  should  have  shown  that  the  integral  part  of  the 
root  of  f(x)  =  0  was  478. 

On  solving  numerical  equations.  If  asked  to  find  uil  the  real  846 
roots  of  a  given  numerical  equation  /(a*)  =  0,  it  is  best,  at 
least  when  the  coefficients  are  rational  numbers,  to  search  first 
for  rational  roots  by  the  method  of  §  802.  This  process 
will  yield  a  depressed  equation  </)(a;)=  0  whose  real  roots,  if 
any,  are  irrational.  We  locate  these  roots  by  the  method  of 
§§  833,  844,  845,  and  then  find  their  approximate  values  by 
Horner's  method. 

It  may  be  added  that  a  fractional  root  may  also  be  found 
by  Horner's  method,  exactly  when  the  denominator  involves 
only  the  factors  2  and  5,  approximately  in  other  cases. 

EXERCISE   LXXV 

Compute  the  roots  indicated  below  to  the  sixth  decimal  figure. 

1.  x^  +  X  —  3  =  0 ;  root  between  1  and  2. 

2.  x^!  +  2  X  -  20  =  0  ;  root  between  2  and  3. 

3.  x''  +  6  x2  +  10  X  -  2  =  0  ;  root  between  0  and  1. 

4.  3  x3  +  5  X  -  40  =  0  ;  root  between  2  and  3. 

6.    x3  +  10  x2  +  8  X  -  120  :=  0 ;  root  between  2  and  3. 


460  A    COLLEGE   ALGEBRA 

6.  2x3  —  x2  —  9x  +  l  =  0;  root  between  —  1  and  —  2. 

7.  x3  +  x2  —  5  X  —  1  =  0 ;  root  between  1  and  2. 

8.  x3  -  2  x2  -  23  X  +  70  =  0 ;  root  between  -  5  and  -  6. 

9.  X*  —  10  x2  —  4  X  +  8  =  0  ;  root  between  3  and  4. 

10.  x*  +  6  x3  +  12  x2  -  11  X  -  41  =  0  ;  root  between  -  2  and  -  3. 

11.  x3  -  3  x2  -  4  X  +  13  =  0  ;  two  roots  between  2  and  3. 

Find  to  the  third  decimal  figure  all  the  roots  of  the  following  equations 

12.  x3  -  3x2 -4x  + 10  =  0.  13.    x3  +  x2-2x-  1=0. 

14.    x3  -  3  X  +  1  =  0.  15.    X*  +  5  x3  +  x2  -  13  X  -  7  =  0. 

16.  By  applying  Horner's  method  to  the  equation  x'  —  17  =  0  compute 
Vl7  to  the  fourth  decimal  figure. 

17.  By  the  same  method  compute  2  Va  and  V87  each  to  the  third 
decimal  figure. 

18.  By  aid  of  §  845  and  Horner's  method  find  to  the  second  decimal 
figure  the  real  root  of  x^  +  x2  -  2500  =  0. 

19.  By  aid  of  §  844  locate  the  roots  of  x^  +  5  x2  -  6  x  +  1  =  0. 

20.  Find  all  the  roots  of  3  x^  +  x«  -  14  x^  -  x2  +  9  x  -  2  =  0. 

TAYLOR'S    THEOREM.     MULTIPLE   ROOTS 

847  Derivatives.  Multiply  any  monomial  of  the  form  ax"  by  n, 
the  exponent  of  x,  and  then  diminish  that  exponent  by  1.  We 
obtain  nax'^~'^,  which  is  called  the  derivative  of  ax",  or,  more 
precisely,  its  derivative  with  respect  to  x.  In  particular,  the 
derivative  of  a  constant  a,  that  is  ax",  is  0. 

The  sum  of  the  derivatives  of  the  terms  of  a  polynomial  f{x) 
is  called  the  derivative  of  f{x),  or,  more  precisely,  its  first 
derivative,  and  is  represented  by /'(a;). 

The  derivative  oi  f  (x)  is  called  the  second  derivative  of 
f{x),  and  is  represented  by  f"{x),  and  so  on. 

Evidently  every  polynomial  f{x)  of  the  nth  degree  has  a 
series  of  »  derivatives,  the  last  of  vvhich, /("^(x),  is  a  constant. 


THEORY   OF   EQUATIONS  461 


Thus,  if 

/(•^) 

=  3x4- 

-8x3  +  4x2- 

■x  +  4, 

we  have 

r{x): 

=  12x3- 

-24x2  +  8x- 

-1, 

f"{^) 

=  36x2 

-48X  +  8, 

r'{x) 

=  72x- 

-48, 

/""(X): 

=  72. 

All  the  subsequent  deriv 

atives  are  0. 

Observe  that  the  second,  third,  •  •  •  derivatives  of  f(x)  are 
the  first,  second,  ••■  derivatives  oif'{x). 

Taylor's  theorem.     If  in  f(x)  =  «yar"  +  a^x"'^  +  ■•  +  «„  we     848 
replace  x  by  x  +  h,  we  obtain 

/(.r  -f  h)  =  a,{x  -f  A)"  +  «!  (a-  +  //)'■'-'  +  •••  +  a„. 

By  expanding  (x  +  //)",  (x  -\-  hy~'^,  and  so  on,  by  the  bino- 
mial theorem  and  then  collecting  terms,  we  can  reduce  this 
expression  to  the  form  of  a  polynomial  in  It.  We  shall  show 
that  the  result  will  be 

f{x  +  h)  =f(x)  +f'(x)  ^  +f"(x)  ~  +  ...  +/«(.•)  ^,    (I) 

where /'(.r),/"(.r),  .. .  are  the  successive  derivatives  of /(a-). 
This  identity  is  called  TayJor^s  theorem. 

For  when  the  result  of  expanding  (x  +  //)"'  by  the  binomial 
theorem,  §  5G1,  is  multiplied  by  a  constant  a  and  written  in 
the  form 

a  (x  +  //)'"  =  ax'"  -f  7nax'"  ~^  •  A  +  m  (tn  —  1)  nx'"    -  ■  — 

+  vi{m  -  l){m-2)ax-'-^ "  |l  +  " " "' 
each  of  the  coefficients 

max"~\    vi(j)i  —  l)ax'"~-,   7n  (jn  —  l)(w  —  2)ax^~^,  ••• 

is  the  derivatioe  of  the  one  which  immediately  precedes  it. 
Hence,  if  we  arrange  the  expansion  of  each  term  of 

/(a-  +  //)  =a,{x  +  hy  +  a,  (x  +  A)""'  +  •••  +  «„ 

in  this  form,  the  sum  of  the  leading  terms  in  these  several 
expansions  will  hef(x);  the  sum  of  the  second  terms  will  be 


462 


A   COLLEGE   ALGEBRA 


849 


850 


h  times  the  sum  of  the  derivatives  of  the  leading  terms,  or 
f'(x)h;  the  sum  of  the  third  terms  will  be  A^/2!  times  the 
sum  of  the  derivatives  of  the  second  terms,  or /"(a;)/i^/2! ; 
and  so  on.     In  other  words,  we  shall  have 


/(.r  +  h)=f{x)+f'{x)h  +/"(x)  -  + 


+r'\-)-,- 


Thus,  if         /(x)  =  aox3  +  a-^xT-  +  OgX  +  a^, 
we  have     /(x  +  h)  =  ao (x  +  h)^  +  ai (x  +  h)'^  +  a^  (x  +  ^i)  +  as 


+  3aox2 

/i  +  6aoX 

+   2aix 

+    2ai 

+        ao 

=      aoX-*   +  3  OoX-  /i  +  0  aox h  6  ao ; 

+  aix2 
+  a2X 
+     as 

=  f{x)  +  r  {z)h  +  f"  {x)h?  /2\  +  f"  [x)hy^\. 
Since    x  =  a  -\-  {x  —  a),    we    have  f(x)  =  /[a  +  (a;  —  a)  ]. 
Hence  we  may  obtain  the  expression  for  f(x)  in  powers  of 
X  —  a,  %  423,  by  merely  replacing  x  by  a,  and  h  hy  x  —  a,  in 
the  identity  (I).     The  result  is 

f{x)=f{a)+f'{a){x-a) 


+  /"(«) 


(x  -  rt)2 


+  f"\a) 


(^  -  «)^ 


(11) 


Example.     Express  x"  —  1  in  powers  of  x  —  1. 
We  have       f{x)  =  x^  -  1,  /'  (x)  =  3  x^  f"  (x)  =  0  x,  /'"  (x)  =  6. 
Hence  /(I)  =  0, /'(I)  =  .3.  /"(l)/2  =  3, /'"(l)/3  !  =  L 

Therefore  x^  -  1  =  3  (x  -  1)  +  3  (x  -  1)2  +  (x  -  1)«. 

Multiple  roots.  The  first,  second,  •  •  •  derivatives  off'(x)  are 
the  second,  third,  -•■  derivatives  oi  f(x).  Hence,  §849,  the 
expressions  for  a  polynomial /(x)  and  its  first  derivative /' (aj) 
in  terms  of  cc  —  a  are 

/(^)  =/(«)  +/'(«)  (-^  -  «)  +/"(«)  (X  -  ay/2  !  +  •••,       (1) 
f'(x)=^f'(a)+f"(a){x  -  a)+f"'{a)(x  -  a)V2!  +  ....   (2) 

lff(x)  is  divisible  by  a-  —  a  but  not  by  (x  —  ay,  it  follows 
from  (1)  that  /(a)  =  0  but  /'  («)  ^  0,  and  therefore  from  (2) 


THEORY   OF   EQUATIONS  463 

that/' (a-)  is  not  divisible  hy  x  —  a.  Again,  if /(sc)  is  divisible 
by  (x  —  ay  but  not  by  (x  —  af,  it  follows  from  (1)  that 
/(a)=/'(a)=0  hut  f"(a)^0,  and  therefore  from  (2)  that 
f'{x)  is  divisible  by  a;  —  a  but -not  by  (x  —  ay.  And  in 
general,  if /(x)  is  divisible  by  (x  —  ay  but  not  by  (x  —  a)''  +  S 
it  follows  from  (1)  that  /(«)  =f'(a)  =  •  •  •/^'"-^^  (a)  =  0  but 
/('''(a)^0,  and  therefore  from  (2)  that  /'(a-)  is  divisible  by 
(x  —  ay~^  but  not  by  (x  —  ay. 

Therefore,  by  §  800,  we  have  the  following  theorem. 

Theorem.    A  simple  root  of  f  (x)  =  0  is  not  a  root  off  (x)  =  0  ,•     851 
but  a  double  root  of  f  (x)  =  0  is  a  simple  root  ofi'(x.)  =  0,  and, 
in  general,  a  multiple  root  of  order  r  of  f  (x)  =  0  is  a  root  of 
order  X-  1  o/f'(x)=0. 

Thus,  the  roots  of /(x)  =  x^  -  x2  ~  8  a;  +  12  =  0  are  2,  2,  -  3,  and  the 
roots  of/'(x)  =  3x2-  2x  -  8  =  0  are  2,  -  4/3. 

We  therefore  have  the  following  method  for  discovering  the  852 
multiple  roots  of  f{x)  =  0,  if  there  be  any.  Seek  the  highest 
common  factor  of  f(x)  and/'(a;)  by  the  method  of  §  465.  If 
we  thus  find  that  f{x)  and/'(a:;)  are  prime  to  one  another, 
f{x)  =  0  has  simple  roots  only.  But  if  we  find  that/(a;)  and 
f'{x)  have  the  highest  common  factor  ^(x),  then  every  simple 
root  of  <l>  (x)  =  0  is  a  double  root  of /(x)  =  0,  every  double  root 
of  <^  (.r)  =  0  is  a  triple  root  off(x)  =  0,  and  so  on.  For,  §  850, 
if  c/)  (,/•)  is  divisible  by  {x  —  ay,  then  /'  (x)  is  divisible  by 
(x  —  ay  and  f{x)  by  (x  —  a)''  +  ^ 

Observe  that  if  the  quotient /(a-)/ <^(a')  be  F(x),  the  roots 
of  F{x)  =  0  are  those  of  f(x)  =  0,  each  counted  once. 

Example.     Find  the  multiple  roots,  if  any,  of  the  equation 
/(x)  =  x5  -  X*  -  5 xs  +  x2  +  8  X  +  4  =  0. 

Here  f  (x)  =  Sx*  -  4x3  _  i5j;2  4.  2x  +  8,  and  by  §  465  we  find  the 
highest  common  factor  of  f{x)  and  /'  (x)  to  be  0  (x)  =  x^  —  3x  —  2. 

The  roots  of  0(x)  =  0  may  be  found  by  §  802  and  are  -  1,  -  1,  2. 
Hence /(x)  =  0  has  the  triple  root  —  1  and  the  double  root  2,  that  is,  its 
roots  are  -  1,  —  1,  —  1,  2,  2. 


464  A   COLLEGE    ALGEBRA 

Observe  that  f{x)  =  (x  +  l)^  (x  -  2)2, 

that  /'(x)  =  (X  +  1)2 (X  -  2)  (5x  -  4), 

and  that  F(x)  =f{x)/<p{x)  =  (x  +  1)  (x  -  2). 

853  We  may  add  that  if  any  ttvo  equations /(x)  =  0  and  i/^  (a-)  =  0 
have  a  root  in  common,  it  may  be  discovered  by  finding  the 
highest  common  factor  off(x)  and  if/(x). 

Example.  Solve/(x)  =  x*  —  x^  —  3  x^  +  4  x  —  4  =  0,  having  given  that 
one  of  Its  roots  Is  the  negative  of  another  of  its  roots. 

Evidently  the  two  roots  mentioned  are  common  to  /(x)  =  0  and  the 
equation  /( —  x)  =  x*  +  X*  —  3x2  —  4  x  —  4  ==  0,  and  may  therefore  be 
obtained  by  finding  the  highest  common  factor  of /(x)  and/(— x). 

By  §  465  we  find  this  highest  common  factor  to  be  x2  —  4.  Hence 
the  roots  mentioned  are  2,  —  2.  Dividing /(x)  by  x2  —  4  and  solving  the 
resulting  depressed  equation  x^  —  x  +  1  =  0,  we  find  that  the  other  two 
roots  of  /(x)  =  0  are  (1  ±  i  Vs)  /2. 

EXERCISE  LXXVI 

1.  Find  the  first,  second,  •  •  •  derivatives  of  2  x*  —  4  x*  +  x2  —  20 x. 

2.  Given /(x)  =  x*  -  2x3  +  1,  find/(x  +  h)  by  Taylor's  theorem. 

3.  Using  the  formula  §  849,  (II),  express  (1)  x*  +  x2  +  1  in  powers  of 
X  +  1  ;  (2)  x5  -  32  in  powers  of  x  -  2  ;  (3)  (x'^  +  1)  /  (x2  +  1)  in  terms 
of  X  —  L 

4.  The  following  equations  have  multiple  roots.     Solve  them. 

(1)  x"  _  3a;  _  2  =  0.  (2)  9x3  +  i2x^  -  llx  +  2  =  0. 

(3)4x4  +  12x2  +  9  =  0.  (4)  x*-4x3  +  8x  +  4  =  0. 

(5)  2  X*  -  12  x3  +  19  x2  -  G  X  +  9  =:  0. 

(6)  x5  -  x3  -  4x2  -  3x  -  2  =  0. 

(7)  x*  -  2  x3  -  x2  -  4  X  +  12  =  0. 

(8)  x5  -  x*  -  2  x3  +  2  x2  +  X  -  1  =  0. 

(9)  3x5-2x*  +  6x3-4x2  +  3x-  2  =  0. 

5.  Show  that  x"  —  a"  =  0  cannot  have  a  multiple  root. 

6.  If  the  equation  x^  —  12x  +  a=0  has  a  double  root,  find  a. 

7.  Determine  a  and  b  so  that  3  x^  +  ax2  +  x  +  6  =  0  may  have  a  triple 
root,  and  find  this  root. 


THEORY    OF    EQUATIONS  465 

8.  Show  that  x*  +  qx~  +  s  =  0  cannot  have  a  triple  root. 

9.  Fhid  the  condition  that  x^  —  px-  +  r  =  0  may  have  a  double  root. 

10.  What  is  the  form  of /(x)  if  it  is  exactly  divisible  by /'(x)  ? 

11.  The  equations  x*  +  x^  +  2  x-  +  x  +  1  =  0  and  x^  +  x^-x-l^O 
have  roots  in  common.     Solve  both  eciuations. 

12.  The  equation  x^  —  20x  —  16  =  0  has  a  root  which  is  twice  one  of 
the  roots  of  x'  —  x-  —  3x  —  1  =  0.     Solve  both  of  these  equations. 

13.  Show  that  if  a  cubic  equation  with  rational  coefficients  has  a 
multiple  root,  this  root  must  be  rational. 

14.  Show  that  if  an  equation  of  the  fourth  degree/(x)  =  0  with  rational 
coefficients  has  a  multiple  root,  this  root  must  be  rational  unless /(x)  is  a 
perfect  square. 

15.  Prove  that  if  a  is  a  root  of  f{z)  =  0,  of  order  r,  it  is  a  root  of  all 
the  equations /'(x)  =  0,  /"(x)  =  0,  ■  •  ■ ,  /('■-i)(x)  =  0. 

VARIATION   OF  A   RATIONAL   INTEGRAL   FUNCTION 

Theorem.  Let  f  (x)  denote  a  polynomial  arranged  in  ascend- 
ing powers  of  x,  and  let  b  denote  the  numerical  value  of  its 
leading  coefficient  and  g  that  of  its  numerically  greatest  coeffi- 
cient. The  leading  term  of  f  (x)  vjill  be  nunierically  greater 
than  the  sum  of  the  remaining  terms  for  all  values  ofx  ivhich 
are  numerically  less  than  b/(b  +  g). 

First,  let  f(x)  =  ^o  +  ^i^  +  ^2*^  +  •  •  •>  so  that  b  =  |^o|>  a,nd 
let  x'  denote  the  numerical  value  of  x. 

Then  b-^x  +  b^x'^  +  •  •  •  is  numerically  less  than  (or  equal  to) 

gx'  +  gx'^  -\ OT  g(x'  +  x'^  ^ ),  §  235,  and  therefore  when 

x' <  1,  it  is  less  than  gx' /(I  —  x'),  §  704. 

But  gx' /  (1  —  x')  is  less  than  b  when  x'  <  b/(b  +  g). 

Second,  let f(x)  =  biX  +  b^x^  +  b^x^  -\ ,  so  that  b=\bi\.    We 

then  have  \b^x^  +  b^x^  -\ 1  <  |^>ia;|  when  \b^x  +  b^x^  +  •  •  •  |  <  |^i|, 

that  is,  when  x'  <  b  / (b  +  g),  and  so  on. 

Thus,  if  /{x)  =  5x  +  3x2-9xS  we  have  |3x2  -  9x*|<(5x|  when 
x'<5/(5  +  9),  that  is,  when  x'<6/14. 


466  A   COLLEGE   ALGEBRA 

855  Theorem.  Let  f  (x)  denote  a  polynomial  arranged  in  descend-^ 
ing  poivers  of  x,  and  let  a  denote  the  numerical  value  of  its 
leading  coefficient  and  g  that  of  its  numerically  greatest  coeffi- 
cient. The  leading  term  of  f  (x)  will  he  numerically  greater 
than  the  sum  of  the  remaining  terms  for  all  values  of  x  which 
are  numerically  greater  than  (a  +  g)/a. 

For  let/(a;)  =  a^x""  +  a-^pcr--^  + h  «„,  so  that  a  =  \a^\,  and 

let  ic'  denote  the  numerical  value  of  x. 

We  have  a^""  +  a^x''-^  H h  «„  =  a;"  («n  +  ai/x  H 1-  «"/.i-"). 

Hence  |aoX"|>|aia:;''-*H |-a„|  when  |a„|>  jaj/ic-^ |-«„/ic"|. 

But,  §  854,  |«o|>|«i/a^  -\ h  «„/«"!  when  l/a;'<a/(a  +  y), 

that  is,  when  x'  >  {a  -\-  g)  / a. 

Thus,  if /(x)  =  3x3  +  x2  _  7a;  +  2,  we  have  |3x3|>|x2  -  7x  +  2]  when 
x'>(3  +  7)/3,  that  is,  wlien  x'>  10/3. 

From  this  theorem  it  evidently  follows  that  the  number  (a  +  g)/a  is 
greater  than  the  absolute  or  numerical  value  of  any  root  of  the  equation 
f(x)  =  0,  whether  the  root  be  real  or  imaginary. 

856  Theorem.  If  2.  is  a  root  oft(x)  =  0,  the  values  of  f  (x)  and 
f'(x)  haoe  contrary  signs  when  x  is  slightly  less  than  a,  ayid  the 
same  sign  when  x  is  slightly  greater  than  a. 

For  express  /(.r)  and/'(a-)  in  powers  of  .r  -  a,  §  849,  and 
then  divide  the  first  expression  by  the  second.  When  a  is  a 
simple  root,  so  that  /(a)  =  0  but/'(«)  -^  0,  the  result  may  be 
reduced  to  the  form 

fM,=U_  „^  /'(^)+.r(^0(^-«)/2!  +  --- 
f'{x)       ^  ^      /'(a)+/"(a)(x-a)  +  ... 

The  numerator  and  denominator  of  the  fraction  on  the  right 
are  polynomials  in  cc  —  a.  Hence  for  all  values  oi  x  —  a  which 
are  small  enough  to  meet  the  requirements  of  §  854  their  signs 
will  be  those  of  their  cominon  leading  term  /'(a),  and  the 
fraction  itself  will  be  positive.  The  sign  oi  f{x)/f'{x)  will 
then  be  the  same  as  that  of  ic  —  a  and  therefore  minus  or  plus 
according  as  x  <  a  or  x  >  a.     But  when  the  sign  of  f{x)  //'  {x) 


THEORY    OF    EQUATIONS  467 

is  minus,/(.r)  and/' (a:-)  have  contrary  signs,  and. when  the  sign 
of  f(x)/f'  (x)  is  plus, /(ic)  and /'(a;)  have  the  same  sign. 

Wlien  a  is  a  multiple  root  of  order  r  we  have,  §  850, 
f(x)  _  f^''\a) / r\  +  terms  involving  (x  —  a) 

f'{x)      ^  /^'H'*)/(*'  —  1)!  +  terjns  involving  {x  —  a) 

from  which  the  theorem  follows  by  the  same  reasoning  as  when 
a  is  a  simple  root. 

RoUe's  theorem.  Between  two  consecutive  roots  ofi(x)—0 
there  is  always  a.  root  o/f'(x)  =  0. 

For  let  Py  and  /So  be  the  roots  in  question,  and  let  c  denote 
a  number  slightly  greater  than  ySi  and  d  a  number  slightly  less 
than  /5o,  so  that  ^i  <  c  <  d  <  ft.. 

Then  f'(c)  has  the  same  sign  as  f(c),  §  856,  and  /(c)  has 
the  same  sign  as  f(d),  §  834 ;  but  f(d)  has  a  different  sign 
from  that  of  /'  (d),  §  856.  Hence  /'  (c)  and  /'  (d)  have  contrary 
signs.  Therefore  a  root  of /'(a')=0  lies  between  c  and  d, 
that  is,  between  fi^  and  ^.,,  §  833. 

Thus,  if  fix)  =  x2  -  3  a:  +  2  =  0,  then  f'{x)  =  0  is  2  x  -  3  =  0.  The 
roots  of  /(x)  =  0  are  1  and  2,  the  root  of  f'(x)  =  0  is  3/2,  and  3/2  lies 
between  1  and  2. 

Example.  Prove  that  f{x)  =  x^  +  x2  -  10  a;  +  9  =  0  has  two  roots 
between  1  and  2.     (Compare  §844,  Ex.) 

Since  /(I)  =  1  and/(2)  =  1,  there  are  two  roots  or  none  between  1 
and  2.  If  there  are  two  roots,  /'  (x)  =  0  must  also  have  a  root  between 
1  and  2,  and  this  root  must  lie  between  the  two  roots  of /(x)  =  0. 

But  /'(x)  =  3x'- +  2x  —  10  =  0  has  a  root  between  1  and  2,  for 
/'  (1)  =  —  5  and  /'  (2)  =  fl.  Solving,  we  find  that  this  root  is  1.5  approxi- 
mately. Moreover /(1. 5)  =  —  .375  is  minus.  Therefore,  since  both /(I) 
and  f{2)  are  plus,  /(x)  =  0  has  two  roots  between  1  and  2,  namely,  one 
between  1  and  1.5,  and  another  between  1.5  and  2. 

Theorem.  If  the  variable  x  is  increasing,  then,  as  it  passes 
through  the  value  a,  the  value  o/f  (x)  is  increasing  if  t' (a,)  >  0, 
but  decreasing  //f'(a)  <  0. 

/f  f'(a)=:  0  but  f"(a)  ^  0,  f  (a)  is  a  maximum  value  ofi(x) 
when  f''(a)  <.  0,  a  minimum  value  when  f"(a)  >  0. 


468  A    COLLEGE   ALGEBRA 

For  by  §  849  we  have 

f{x)-f{a)=f\a){x  -  a)+f"{a){x  -  ay/2  !  +  ••.. 

The  second  member  is  a  polynomial  in  x  —  a,  and  for  all 
values  of  x  which  make  x  —  a  small  enough  numerically  to 
meet  the  requirement  of  §  854,  the  leading  term  will  control  the 
sign  of  the  entire  expression  and  therefore  that  of /(a-)— /(a). 
We  shall  suppose  x  restricted  to  such  values.     Then 

1.  If  f'{a)  >  0,  fia)  {x  -  a),  and  therefore  f{x)  -f{a), 
has  the  same  sign  as  (x  —  a).  Therefore,  since  x  —  a  changes 
from  minus  to  plus  as  x  passfct,  through  a,  the  same  is  true  of 
f(x)—f{a),  that  is, /(x)  is  then  increasing  from  a  value  less 
than/(«)  to  a  value  greater  than /(re). 

2.  If  f'{a)  <  0,  /'(«)  {x  -  a)  and  (x  -  a)  have  contrary 
signs.  Hence,  reasoning  as  in  1,  we  conclude  that  f(x)  is 
decreasing  as  x  passes  through  a. 

3.  If/'(«)=0  but/"(a)^0,  the  sign  oif(x)-f(a)  is 
that  of  /"(«)(.!■  — r^y-/ 2  and  therefore  that  off"(a);  for 
(x  —  a)'^  is  positive  whether  x  <a  or  x  >  a.  Hence  when 
/"{")  <  0,  we  have/(u:-)  </(«)  just  before  x  reaches  a  and  also 
just  after  x  passes  a,  which  proves  that  f(")  is  a  viax'unuvi 
value  oif{x),  §  639. 

And  vn  the  same  manner  when/"(rt)  >  0,  we  may  show  that 
f{a)  is  a  minlviiDii  value  of /(a*). 

It  may  be  added  that  if  /"  {a)  =  0  but  /'"  (a)  ^  0,  /(a) 
is  not  a  maximum  or  minimum  value  of  f{x)  (see  §  859, 
Ex.  2).  And,  in  general,  if  all  the  derivatives  from  the  hrst 
to  the  rth,  but  not  the  {r  +  l)th,  vanish  when  x  =  a,  f(a) 
is  a  maximum  or  minimum  when  r  is  odd,  but  not  when  r  is 
even. 

Example.  Is/(x)  =  x^  —  Gx^  +  9x  —  1  increasing  or  decreasing  as  x, 
increasing,  passes  through  the  value  2  ?  Find  the  maxinuun  and  mininuim 
values  of/(x). 

We  find/'  (x)  =  3 x2  -  12 X  +  9  =--  3  (X  -  1)  (x  -  3).  Hence/'  (2)  =  -  3 
is  negative.     Therefore /(x)  is  decreasing  as  x  passes  through  2, 


THEORY   OF    EQUATIONS 


469 


We  have /'  (x)  =  0  when  x  =  1  and  when  x  =  3.  Moreover/"  (x)  =  6  x  — 12, 
and  therefore  /"  (1)  =  -  6  is  negative  and  /"  (3)  =:  6  is  positive.  Hence 
/(I)  =  3  is  a  maximum  value  of /(x),  and  /(3)=  — Lis  a  minimum 
value. 

Variation  of  f  (x).  Let  us  now  consider  how  the  value  of  a 
polynomial  f(x)  with  real  coefficients  varies  when  x  varies 
continuously,  §  214,  from  —  co  to  +  oc. 

Example  1.     Discuss  the  variation  of /(x)  =  x^  —  2x2  —  x  +  2. 

The  roots  of /(x)  =  0  are  -  1,  1,  2,  and  /(x)  =  (x  +  1)  (x  -  1)  (x  -  2). 

Hence,  when  x  =  —  co,  /(x)  =  —  co  •  when  x  is  between  -co  and  —  1, 
/(x)  is  negative  ;  when  x  =  —  1,  /(x)  >-  ^J ;  when  x  is  between  —  1  and  1, 
/(x)  is  positive  ;  when  x  =  1,  /(x)  —  0 ;  wlien  x  is  between  1  and  2,  /(x) 
is  negative  ;  when  x  =  2,  /(x)  =  0 ;  when  x  is  between  2  and  cc,  /(x)  is 
positive  ;  when  x  =  cc,  /(x)  =  co. 

The  roots  of /' (x)  =  3x2  -  4x  -  1  =  0  are  (2  ±  V7)/3,  or  -  .2  and  1.5 
approximately.  When  x  <  (2  -  V?)  /  3  and  when  x  >  (2  +y7)  /3,  /'  (x) 
is  positive,  but  when  x  is  between  (2  —  V7)/3  and  (2  -f  V7)/3,  /'(x)  is 
negative. 

Therefore,  §  858,  /(x)  is  continually  increasing  as  x  varies  from  —  co 
to  (2  —  V7)/3,  is  continually  decreasing  as  x  varies  from  (2  —  V7)/3 
to  (2+V7)/3,  and  is  again  continually  increasing  as  x  varies  from 
(2  +  V7)/3  to  00. 

It  follows  from  this,  §  639,  that  /(x)  has  a  maximum  value  when 
X  =  (2  —  Vt)  /3,  and  a  minimum  value  when  x  =  (2  +  V?)  /3.  This  is  in 
agreement  with  §  858,  for/"  (x)  =  6  x  —  4  is  negative  when  x  =  (2  —  V7)/3, 
and  positive  when  x  =  (2  +  V7)/3. 

The  variation  of  f(x)  will  be  ex- 
hibited to  the  eye  if  we  put  y  =:f{x) 
and  then  construct  the  graph  of  this 
equation  by  the  method  of  §  389. 
We  thus  obtain  the  curve  indicated 
in  the  accompanying  figure.  The 
points  A,  B,  C  at  which  the  curve 
cuts  the  X-axis  are  the  graphs  of  the 
roots  -  1,  1,  2  of /(x)  =  0.  The  por- 
tions of  the  curve  above  the  x-axis 
correspond  to  positive  values  of /(x), 
those  below  to  negative  values.  The  uppermost  point  on  the  curve  between 
A  and  B  corresponds  to  the  maximum  value  of /(x),  the  lowermost  point 
between  B  and  C  to  the  minimum  value. 


859 


470 


A  COLLEGE  ALGEBRA 


860 


As  X  varies  from  —  »  to  x  the  corresponding  point  on  the  curve  moves 
from  an  infinite  distance  belov?  the  x-axis  upward  and  to  the  right  through 
A  to  the  maximum  point,  then  downward  through  B  to  the  minimum  point, 
then  upward  again  through  C  to  an  infinite  distance  above  the  cc-axis. 

If  we  were  gradually  to  increase  the  absolute  term  of /(x),  the  graph 
of  y  =f{x)  would  be  shifted  vertically  upward  and  the  points  B  and  C 
would  at  first  approach  coincidence  in  a  point  of  tangency  and  then  dis- 
appear. The  corresponding  roots  of /(j)  =  0  would  at  first  become  equal 
and  then  imaginary. 

Example  2.     Discuss  the  variation  of /(x)  =  x*  —  2  x^  +  2  x  —  1. 
The  roots  of /(x)  =  0  are  -  1,  1,  1,  1,  and/(x)  =  (x  +  1)  (x  -  1)-^. 
Hence  when  x  =  ±  oo,  /(x)  =  co ;  when  x  =  ±  1 ,  f{3^)  =  0  ;  when  x  <  —  1 
and  when  x>l,  f(x)  is  positive;  when  x  is  between  —  1  and  1,  /(x)  is 
negative. 

Here  /'  (x)  =  4  x^  -  6  x2  +  2  =  2  (2  x  +  1)  (x  -  1)2,  and  the  roots  of 
/'(x)  =  0  are  —1/2,  1,  1.  When  x<  — l/2,/'(x)  is  negative;  when  x 
is  between  —  1/2  and  1  and  also  when  x  >  1,  /'  (x)  is  positive.  Therefore, 
§858,  f(x)  is  continually  decreasing  as  x  varies  from  —  oo  to  —1/2, 
and  continually  increasing  as  x  var  es  from  —  1/2  to  1  and  from  1  to  oo. 
Hence  f(x)  has  a  minimum  value  when 
X  =  —  1/2,  but  it  has  neither  a  maxi- 
mum nor  a  minimum  value  when  x  =  1. 
This  is  in  agreement  with  §  858. 
For  /"  (x)  =  12  x2  -  12  X  =  12  X  (X  -  1) 
and  /'"(x)  =  24x  —  12.  Hence  when 
X  =  —  1  /2,  /"  (x)  is  positive  ;  but  when 
X  =  1 ,  /"  (X)  =  0  and  /'"  (x)  ^  0. 

The  graph  of  y  =f{x)  has  the  form 
indicated  in  the  accompanying  figure. 
The  point  A  where  the  curve  merely 
cuts  the  X-axis  corresponds  to  the  root 
—  1  of /(x)  =  0,  and  the  point  B  where 
the  curve  both  touches  and  crosses  the 
X-axis  corresponds  to  the  triple  root  1. 
The  lowermost  point  of  the  curve  corresponds  to  the  minimum  value  of 
/(x).     Its  coordinates  are  —1/2,  —27/16. 

As  in  these  examples,  so  in  general,  iif(x)  is  of  odd  degree, 
its  leading  coefficient  being  positive,  when  x  varies  from  —  oo 
to  -f-  cc,f(x)  increases  from  —  oo  to  the  first  maximum  value, 
then  decreases  to  the  first  minimum  value,  and  so  on,  and  finally 


THEORY   OF   EQUATIONS  471 

increases  from  the  last  miniuiuin  value  to  +  oo.  It  is  possible, 
however,  that  there  are  no  maximum  or  minimum  values,  for 
the  equation /'(x)  =  0,  being  of  even  degree,  may  have  no  real 
root.  The  graph  of  y  =/(a-)  extends  from  an  infinite  distance 
below  the  a;-axis  to  an  infinite  distance  above  the  cc-axis.  It 
crosses  the  a;-axis  an  odd  number  of  times,  —  once  at  least. 

On  the  other  hand,  if  f{x)  is  of  even  degree,  f{x)  begins  by- 
decreasing  from  +  CO  to  the  first  minimum  value  and  ends 
by  increasing  from  the  last  minimum  value  to  +  oo.  In  this 
case  the  graph  of  y  —f{x)  need  not  cross  the  a;-axis  at  all.  If 
it  does  cross  the  axis,  it  crosses  an  even  number  of  times. 

In  most  cases  we  can  obtain  a  sufficiently  accurate  represen- 
tation of  the  graph  of  y  =f(x)  by  the  method  of  §  889,  which 
consists  in  assigning  a  series  of  values  to  cc,  computing  the 
corresponding  values  of  y,  plotting  the  pairs  of  values  of  x,  y 
thus  found,  and  passing  a  "  smooth  "  curve  through  all  these 
points.  Such  a  curve  will  indicate  roughly  where  the  true 
graph  crosses  the  cc-axis  and  where  its  maximum  and  minimum 
points  lie.  But  to  obtain  the  points  of  crossing  with  exactness, 
we  must  solve  the  equation  f{x)  =  0  ;  and  to  obtain  the  actual 
positions  of  the  maximum  and  minimum  points,  we  must  solve 
the  equation  /'  (x)  =  0.  To  every  multiple  root  of  f(x)  =  0 
there  corresponds  a  point  of  tangency  of  the  graph  with  the 
X-axis.  If  the  order  of  the  multiple  root  is  odd,  the  graph 
also  crosses  the  x-axis  at  this  point. 

EXERCISE   LXXVII 

1.  Discuss  the  variation  of  /(«)  =  (x  +  1)  (a;  -  2)2  =  x^  -  3x2  +  4, 
finding  its  maximum  and  minimum  values  if  any,  and  draw  the  graph  of 

y=f{^). 

2.  Treat  in  a  similar  manner  each  of  the  following  functions. 
(I)2x2-x  +  l.  (2)  (x  +  l)(x-2)(2x-l). 
(3)  x3  -  12x  +  14.                             (4)  x3  -  5x2  +  33;  +  9. 

(5)  x3  -  3  x2  +  5.  (6)   (X  +  1)2  (X  -  2)2. 

(7)  (x2  +  X  +  1)  (X  +  2).  (8)  x(x  -  1)  (X  4  2)  (x  +  3). 


472  A    COLLEGE   ALGEBRA 

3.    Find  the  graphs  of  each  of  the  foHowing  fractional  equations  by 
plotting  the  points  corresponding  to  x  =  —  1,  —  1/2,  0,  1/2,  1,  •  •  •,  4. 

(1),=        ^^^-'^       ■  i2)y=       "(^-^)       ■ 

(X  -  2)  (X  -  3)  (X  -  1)  (X  -  3) 


STURM'S    THEOREM 

361  Sturm  functions.  Let  f(x)  =  0  be  any  equation  which  has 
no  multiple  roots,  and  let/i(a;)  be  the  first  derivative  of /(cr). 

Divide  f{x)  by  fi(x')  and  call  the  quotient  qi,  and  the 
remainder,  Avith  its  sign  changed,  /a  (x). 

Again,  divide  /i  (x)  by  /g  (x),  and  call  the  quotient  qo,  and 
the  remainder,  with  its  sign  changed, /a  (a:). 

And  so  on,  modifying  the  ordinary  process  of  finding  the 
highest  common  factor  off(x)  and/i(a;)  in  this  respect  only: 
tJie  sl{/n  of  each  remainder  is  changed,  and  care  is  taken  to 
make  no  other  chanc/es  of  sign  than  these.. 

Since /(ic)  ==  0  has  no  multiple  roots  and  therefore /(cc)  and 
fi(x)  have  no  common  factor,  §  851,  we  shall  finally  obtain  a 
remainder  which  is  a  constant  different  from  0,  §  465.  Call 
this  remainder,  with  its  sign  changed,  /„,. 

The  sequence  of  functions 

consisting  of  the  given  polynomial,  its  first  derivative,  and 
the  several  remainders  in  order,  each  with  its  sign  changed,  is 
called  a  sequence  of  Sturm,  or  a  sequence  of  Sturm  functio7is. 
862         Relations  among  the  Sturm  functions.     These  functions  are, 
by  definition,  connected  by  the  series  of  identical  equations 

f{x)^q,f(^x)-f{x),  (1) 

M^)^'i.A{^)-M^),  (2) 

f,(x)^q,f,{x)-f,(x),  (3) 

fm  -2^^)=  1,n  -  J.n  - 1  (^)  "  fm-  (m  -  1) 


THEORY    OF    EQUATIONS  473 

From  these  equations  we  conclude  that 

1.  Two  consecutive  functions  cannot  vanish  for  the  same 
value  of  X. 

Thus,  if  both/i(a;)  and/2(a;)  vanish  when  a:;  =  c,  it  follows 
from  (2)  that  fz{x)  also  vanishes;  therefore,  from  (3),  that 
fi{x)  vanishes;  and  therefore,  finally,  that /„,  is  0.  But  this 
is  contrary  to  hypothesis. 

2.  When  for  a  certain  value  of  x  one  of  the  intermediate 
functions  f^  (x),  fg  (x),  •  •  •,  fm-i  (a;)  vanishes,  the  functions  tvhich 
immediately  precede  and  follow  it  have  opposite  signs. 

Thus,  if  /a  (c)  =  0, 

it  follows  from  (2)  that      /i  (c)  =  -  /s  (c). 

Sturm's  theorem.    Let  a  and  b  he  any  two  real  numbers  neither     863 

of  which  is  a  root  o/f  (x)=  0. 

The  difference  betiveen  the  number  of  variations  of  sign  in 

the  sequence  r  /  \     *  /  x     ^  /  \  * 

f(a),  fi(a),  f2(a),  ••-,  f„ 

and  that  in  the  sequence 

f(b),  f,(b),  f,(b),   ...,  f„ 

is  the  number  of  roots  of  i(yi)  =  0  which  lie  between  a  and  b. 

To  fix  the  ideas,  suppose  a  <  b,  and  suppose  x  to  vary  con- 
tinuously from  ft  to  ft,  §  214. 

As  X  varies  from  a  to  b,  the  sign  of  the  constant  /„,  remains 
unchanged,  and  the  only  changes  which  are  possible  in  the 
signs  of  the  remaining  functions,  and  therefore  in  the  number 
of  variations  of  ^n  in  the  sequence 

f(x),A(x),f,(x),  ...,/„, 
are  such  as  may  occur  when  x  passes  through  roots  of  the 
equations  f(x)  =  0,  fi(x)  =  0,  and  so  on,  §  835.     But 

1.  The  number  of  variations  in  the  sequence  is  neither 
increased  nor  diminished  when  any  function  except  the  firsty 
f  (x),  changes  its  sign. 


474  A   COLLEGE   ALGEBRA 

Suppose,  for  instance,  that  c  is  a  root  of  /,  (x)  =  0  and  that 
/a  (x)  changes  from  plus  to  minus  as  x  passes  through  c. 

Since  c  is  a  root  of  /g  (x)  =  0,  it  cannot  be  a  root  of  either 
fi  (x)  =  0  or  /a  (x)  =  0,  §  862.  And  if  we  take  a  positive 
number  7^  so  jmall  that  jig_  root  of  fiJ^X^^  ^^'  o^  Ai^l^  ^ 
lies  between  c  —Jj  and  c,  or  between  g.  and  c  +  h,  neither  of 
the  functions  fi(x)  or  /^(x)  will  change  its  sign  as  x  varies 
from  c  -  A  to  c  +  h,  §  835. 

But  when  x  =  c,f^{x)  and/3(a:)  have  opposite  signs,  §  862. 
Suppose  that  /i(c)  is  plus;  then  /3(c)  is  minus,  and  we  have 
the  following  scheme  of  the  signs  o^  fi{x),  /^{x),  f^(x)  for 
values  of  x  between  c  —  h  and  c  -\-  h  : 

fiix)    h{x)    fz{x) 
x  =  c  —  h 


c  +  h 


+  + 

+       a 


Eor  all  these  values  of  x,  therefore,  the  signs  of  the  three 

functions 

A(x),  f,(x),  f,(x) 

present  o7ie  variation.  The  only  effect  of  the  change  of  the 
sign  oif^^x),  at  least  in  this  part  of  the  sequence,  is  a  change 
in  the  ^osiVion  of  a  variation. 

And  there  can  be  no  cliange  in  the  number  of  variations 
in  the  part  of  the  sequence  which  follows /g  (ic)  as  x  passes 
through  c.  For  either  no  function  after  fz(x)  will  change  its 
sign  as  x  passes  through  c,  or,  if  it  does,  we  shall  have  the 
case  of  the  function  /j  (x)  over  again,  and  the  only  effect  will 
be  a  change  in  the  position  of  another  variation. 

2.  The  sequence  loses  one  variation  for  each  root  of  i(x)  =  0 
through  which  x  passes  in  vanjing  from  a  to  b. 

For  let  c  be  a  root  oif(x)  =  0. 

The  functions /(a;)  and/i(ir)  have  opposite  signs  for  values 
of  X  which  are  slightly  less  than  c,  and  the  same  sign  for 
values  of  x  which  are  slightly  greater  than  c,  §  856. 


THEORY    OF    EQUATIONS  475 

In  other  words,  the  sequence  has  a  variation  between  f{x) 
and./i(a')  just  before  x  reaches  c,  and  this  variation  is  lost  as 
X  passes  through  c. 

Hence  the  sequence  of  Sturm  functions  never  gains  a  varia- 
tion as  X  varies  from  a  to  b.  But,  on  the  other  hand,  it  loses 
a  variation  each  time  that  x  passes  through  a  root  oi  f(x)  =  0, 
and  then  only. 

Therefore  the  difference  between  the  number  of  variations 

and  in  f(b),  f,{b),  f,{b),   ••.,/„. 

is  the  number  of  single  variations  that  are  lost  as  x  passes 
through  the  roots  of /(x)  =  0  which  lie  between  a  and  b.  In 
other  words,  it  is  the  number  of  these  roots,  as  was  to  be 
demonstrated. 

If  we  apply  the  method  of  §  861  to  an  equatiou  /(x)  =  0  which  has 
multiple  Yooi?.,  we  obtain  a  sequence/ (x), /i  (x),  •  •  •,/,„(x),  (1)  the  last  term 
of  which  is  the  highest  common  factor  of  all  the  terms,  §  4(55.  Divide  all 
the  terms  of  (1)  by  f,„{x).  We  thus  obtain  a  sequence  of  the  form  0(x), 
01  (x),  •  •  • ,  1,  (2)  which,  as  is  easily  shown,  possesses  all  the  properties  on 
which  Sturm's  theorem  depends.  Hence  the  number  of  roots  of  0  (x)  =  0, 
that  is,  §  852,  the  number  of  different  roots  of  f{x)  =  0,  between  a  and  6, 
is  the  difference  between  the  number  of  variations  in  0(a),  0i(a),  •  •  •,  1 
and  in  0(6),  <Pi(b),  •••,  1.  And  this  difference  is  the  same  as  that 
between  the  number  of  variations  in  /(a),  /i(a),  •  ■  -,  fm{a)  and  in  /(6), 
/i  (6)i  •  •  •  ?  fm  (b) ;  for  multiplying  the  sequences  0  (a),  •  •  • ,  1  and  0  (6),  •  •  • ,  1 
'^y  fm(a)  and/,„(5)  respectively  will  not  affect  their  variations. 

Applications  of  Sturm's  theorem.  Sturm's  theorem  enables  864 
one  to  find  exactly  how  many  different  real  roots  a  given 
numerical  equation  has.  It  also  enables  one  to  find  how 
many  of  these  roots  lie  between  any  pair  of  consecutive  inte- 
gers and  therefore  in  every  case  to  solve  the  problem  of  locat- 
ing the  roots.  But  this  method  of  locating  roots  is  very 
laborious  and  is  used  only  when  the  sin;pler  method  of  §  836 
fails. 


476 


A   COLLEGE    ALGEBRA 


Example  1.     Locate  the  roots  of  x^  +  Sx^  —  4x  +  1  =  0. 
Here/(x)  =  x^  +  3x2  -  4x  +  1  aiid/i(x)  =  3x2  ^.  6x  -  4. 
Arranging  the  computation  of  the  remaining  functions  as  in 
have 

1+3-    4+1 

3  +  9-12  +  3 

3  +  6-    4 


i,  3,  we 


3+    6 
6  +  12 


15- 

8 

30- 

16 

30- 

15 

3- 


8  +  3 
6-4 


/2 


14  +  7 
=  2-1 


Hence 

/(x)  =  x3  +  3x2- 
/i(x)  =  3x2  +  6x 
/2(x)  =  2x-l, 

/3  =  L 


3  +  15 

/3  =  1 

Observe  that  the  fz  (x)  and  /s  here  obtained  are  not  the  f^  (x)  and  /s 
defined  in  §  861,  but  these  functions  multiplied  by  positive  constants* 

We  should  have  lessened  the  reckoning  had  we  divided  /i(x)  by 
/2(x)  =  2(x  —  .5)  synthetically.  It  is  often  best  to  use  the  synthetic 
method  in  the  final  division. 

1.  When  X  is  very  great  numerically,  the  sign  of  a  polynomial  is  that 
of  its  term  of  highest  degree,  §  855.     Hence  the  following  table. 

fix)    .h(x)     f.,(x)     ,h 

+ ,  three  variations. 
+ ,  t^^o  variations. 
+ ,  no  variation. 
Therefore /(x)  =  0  has  one  negative 'and  two  positive  roots. 

2.  To  locate  the  positive  roots,  we  substitute  x  =  0,  1,  •  •  •  and  obtain 
f{x)     /i(x)    /o(x)    /3 

0 "+     I     —     I      ^^^     r  + ,  two  variations. 


x  =  l 


+ 


no  variation. 


Hence  the  two  positive  roots  lie  between  0  and  1. 

Since  there  is  but  one  negative  root,  we  know  that  it  can  be  located  by 
ihe  method  of  §  836.     We  thus  find  that  it  lies  between  —  4  and  —  5. 

Example  2.     How  many  real  roots  has  /(x)  =  2x'  —  x^- 2x  +  2  =  0? 

Proceeding  as  in  Ex.  1,  we  find/i(x)=  3x2  —  x  —  1  and/2(x)  =  13x  — 17. 

We  can  find  the  sign  of  /a,  which  alone  concerns  us,  without  dividing 
/i(x)by/2(x). 

For  since /j  (x)  =  13(x  —  17/13),  the  remainder  in  the  division  of /i(x) 
by/2(x)  is/i(17/13).  But  17/13>1,  and  fx{x)  is  positive  when  x>l. 
Hence  /i  (17/13)  is  jiositive,  and  therefore /s  is  negative. 

For  X  =  —  00  the  signs  of  /(x),  /i  (x),  /2(x),  fz  are  — ,  +,—,—;  for 
X  =  00  they  are  +,  +,  +,  — .     Hence  f{x)  —  0  has  but  one  real  root. 


THEORY    OF   EQUATIONS  .477 

The  only  property  of  the  final  function  /„  of  which  any  use 
is  made  in  the  proof  in  §  863  is  that  its  sign  is  constant. 
Hence  if,  when  computing  the  Sturm  functions  of  f(x)  =  0  in 
order  to  find  the  number  of  roots  between  a  and  h,  we  come 
upon  a  function  jf^  (a;)  which  has  the  same  sign  for  all  values 
of  X  between  a  and  b,  we  need  not  compute  the  subsequent 
functions.  For  it  follows  from  the  proof  in  §  863  that  the 
required  number  of  roots  will  be  the  difference  between  the 
number  of  variations  in  f{ci),  ■■•,fp(a)  and  in  f{b),  ■■  ■,fp{b). 

Example  3.     How  many  real  roots  has/(x)  =  x3  +  x2  +  x  +  l  =  0? 

Here  /i  (x)  =  Sx^  +  2  x  +1,  and,  since  22  <  4  •  3,  this  is  positive  for  all 
real  values  of  x,  §§  035,  823.     Hence  we  need  not  compute  /o  (x)  and  /s. 

The  signs  of  /(x),  /i (x)  for  x  =  -  oo  are  -,  + ;  for  x  =  co  they  are 
+  ,  +.     Hence /(x)  =  0  has  one  real  root. 

EXERCISE  LXXVm 

By  aid  of  Sturm's  theorem  find  the  situation  of  the  real  roots  of  the 
following  equations. 

1.    x3  -  G  x2  +  5  X  +  13  =  0.  2.  x3  -  4  x2  -  10  X  +  41  =  0. 

3.    x3  +  5x  +  2  =  0.  4.  x3 +  3x2  +  8x  + 8  =  0. 

5.    x3-x2- 15x4- 28  =  0.  6.  x*  -  4x3- Sx^ +18x  +  20  =  0. 

7.    2x4-3x2  +  3x-l  =  0.  8.  x*  -  8x3  +  19x2 -12x  +  2  =  0. 

9.  x-i  -  12  x2  +  12  X  -  3  =  0.  10.  x*  +  2  x3  -  6  x2  -  8  X  +  9  =  0. 

By  aid  of  Sturm's  theorem  find  the  number  of  the  real  roots  of  each 
of  the  following  equations. 

11.    4x3 -2x- 5  =  0.  12.    x*  +  x3  +  x2  +  x  +  l  =  0. 

13.   X"  +  1  =  0.  14.    X*  -  6  x3  +  x2  +  14  X  -  14  =  0. 

15.  Let  /(x)  =  0  be  an  equation  of  the  ?ith  degree  without  multiple 
roots.  Show  that  the  condition  that  all  the  roots  of /(x)  =  0  be  real  is  that 
there  be  n  +  1  terms  in  its  sequence  of  Sturm  functions /(x),  /i  (x),  •  •  •,/«) 
and  that  the  leading  terms  of  all  these  functions  have  the  same  sign. 

16.  By  aid  of  the  theorem  in  Ex.  15  prove  that  the  condition  that 
all  the  roots  of  the  cubic  x3  +  px  +  5  =  0  be  real  and  unequal  is  that 
4j)3  -I-  27  q2  be  negative. 


478  A   COLLEGE    ALGEBRA 


SYMMETRIC  FUNCTIONS   OF  THE  ROOTS 

865         Theorem  1.    If  the  roots  oft(x)  =  x"  +  bix"-'  -\ h  b„  =  0 

are  /3^,  /3„  ■  ■  ■ ,  fin,  so  that  f  (x)  =  (x  -  13,)  (x  -  ^,)  •  •  •  (x  -  (3,), 

Thus,  suppose  that  7i  =  3,  so  that 

-       f(x)  =  (x-^,)(x-^.;)(x-p,).  (1) 

Substituting  x  +  h  for  x  in  (1),  we  have 
f(x  +  A)  =  [(X  -  /30  +  A]  [(X  -/?,)+  /O  [(^  -  iSa)  +  A].    (2) 

We  can  reduce  each  member  of  (2)  to  the  form  of  a  poly- 
nomial in  h,  the  first  member  by  Taylor's  theorem,  §  848,  the 
second  by  continued  multiplication,  as  in  §  558. 

Since  (2)  is  an  identity,  the  coefficients  of  like  powers  of  /* 
in  the  two  polynomials  thus  obtained  must  be  equal,  §  284. 

But  since  f(x  +  h)=f(x)  +f'(x)h^ ,  §  848,  the  coeffi- 
cient of  h  in  the  first  polynomial  is  f'(x).  In  the  second 
it  is  (x  -  (3,)  {X  -  (3,)  +  (x-  ^83)  (X  -  (30  +  (x-  /30  (x  -  (3,). 

^^""^^     /'  (x)  =  (x-  A)  (X  -  ^3)  +  (x  -  A)  (^  -  A) 

X  -  /3i'^  X  -  ^,      X  -  /3,'  ^  ^ 

since  (x  —  ySg)  (x  —  (3^)  =f{x)/(x  —  (3^,  and  so  on. 

This  reasoning  is  applicable  to  an  equation  of  any  degree  n. 
In  the  general  case  there  are  71  factors  in  the  second  member 
of  (2),  and  when  this  member  is  reduced  to  the  form  of  a 
polynomial  in  h,  the  coefficient  of  h  is  the  sum  of  the  products 
of  the  binomials  x  —  ft,,  x  —  /3.,,  ■  ■  ■ ,  x  —  y8,„  taken  ?i  —  1  at  a 
time,  §  558.  That  one  of  these  products  which  lacks  the  factor 
X  —  fti  may  be  written  f(x)  /  {x  —  /3i),  and  so  on. 


THEORY   OF    EQUATIONS  479 

Thus,  if     /(x)  =  x-^-Gx2  +  llx-G  =  (x-l)(x-2)(x-3), 
we  have        /'  (x)  =  Sx^  —  12  x  +  11 
and  (X  -  2)  (X  -  3)  +  (x  -  3)  (x  -  1)  +  (x  -  1)  (x  -  2)  =  3x2  -  12x  +  n. 

Theorem  2.  Tlte  sums  of  like  powers  of  all  the  roots  of  an 
equation  f  (x)=  0  can  he  expressed  rationallij  in  tervis  of  its 
coefficients. 

Thus,  suppose  that  the  equation  is 

/(a-)  =  ic^  +  b^x""  +  b.x  +  ^3  =  0.  (1) 

Let  a,  /3,  y  denote  the  roots  of  (1)  and  let  s^,  s^,  ■  •  -,  s,.  have 
the  meanings 

si  =  «  +  /?  +  y,     52  =  a2  +  ^-  +  yV  •  ■,  s,  =.«'•  +  /S'-  +  y'-. 
We  are  to  prove  that  s^,  s^,  ■  ■  ■  can  be  expressed  rationally 
in  terms  of  the  coefficients  bi,  h^,  b^. 

1.    By  the  preceding  theorem,  §  865,  we  have 

^    ^        X  —  a        X  —  ft        X  —  y  ^ 

Since  f(x)  is  divisible  by  x  —  a,  x  —  (3,  and  x  —  y,  each  of 
the  fractions  in  (2)  represents  a  polynomial  in  x  which  can 
be  found  by  the  rule  of  §  410.  Applying  this  rule  and  then 
adding  the  results,  we  have 

f(x)  I  {x  -  a)  =  a-^  +  (a  +  ^-i)  a;  +  {a^  +  b^a  +  b^) 
f{x)  /  (;r  -  /?)  =  a:2  +  (^  4-  b,)  X  +  {ft'  +  b.ft  +  b.:) 

f(x)/(x  -  y)=  a;2  ^(y  +  b{)X+(y'  +  b,y  +  b,) 

f  (x)  =3x^^+(s,  +  3b,)x+  (52  +  b,s,  +  3 ^-2)    (3) 
But  by  definition,  §  847,  we  also  have 

f'{x)=3x''  +  2b,x  +  b,.  (4) 

Equating  the  coefficients  of  like  powers  of  x  in  the  two 
expressions  (3)  and  (4)  and  solving  for  s^,  s„,  we  have 

s,  +  3b,  =  2b„  .-.  s,=-b„  (5) 

S2  +  b,si  +  3b2  =  b2,      .-.  s,  =  bl-2  b^.  (6) 


480  A   COLLEGE   ALGEBRA 

2.    From  the  values  thus  found  for  s^  and  s^  we  can  obtain 
the  values  of  s^,  s^,  ■  •■  successively  as  follows  : 
Since  a,  /8,  y  are  the  roots  of  (1),  we  have 

a'  +  bia^  +  ha  +  b^  ^  0,  (7) 

P^  +  b,/3'+b,/3  +  b,  =  0,  (8) 

y'  +  b,y'  +  b,y  +  b,  =  0.  (9) 

Adding  these  identities,  we  obtain 

S3  +  ^1*2  +  Ml  +  3  Z>3  =  0,  (10) 

which  gives  53  rationally  in  terms  of  bi,  b^,  b^,  s^,  s.2  and  there- 
fore, by  (5),  (6),  rationally  in  terms  of  bi,  bo,  b^. 

Next  multiply  the  identities  (7),  (8),  (9)  by  a,  ft,  y  respec- 
tively and  add  the  results.     We  obtain 

Si  +  b,s,  +  b^s,  +  b,s,  =  0,  (11) 

which  by  aid  of  (5),  (6),  (10)  gives  S4  rationally  in  terms  of 
bi,  &2)  K 

And  in  like  manner,  if  we  multiply  (7),  (8),  (9)  respectively 
by  a^,  p-,  y^,  by  a^,  [i^,  y^,  and  so  on,  and  after  each  series  of 
multiplications  add  results,  we  obtain  identities 

S5  +  ^1*4  +  ^2^3  +  ^^3«2   =  0,     Se  +  ^1^5  +  l^"Ji  +  '''3S3  =   0,    •  •  •, 

which  give  S5,  s^,  ■••  rationally  in  terms  of  b^,  b..,  b^. 

By  similar  reasoning  the  theorem  may  be  proved  for  an 
equation  /(;r)  =  0  of  any  degree  n. 

Example  1.     If  a-,  /3,  7  denote  the  roots  oi  x'^  —  2  x"^  +  4  x  +  2  =  0,  find 
Sl/a  =  l/a  +  l//3  +  l/7,  2  l/a2  =  l/a^  +  1/^2  +  1/^2^ 
S  l/a3  =  l/a3  +  I//33  +  1/73. 

Applying  the  transformation  x  —  \/y,  and  dividing  the  transformed 
equation  by  the  coefficient  of  y^,  we  have  y^  -\-2'\p-  —  y  ■\-\  /2  =  ^. 

For  this  equation,  by  substituting  61  =  2,  62  =  —  1,  &3  =  1  /2  in  the 
formulas  (5),  (6),  (10)  above,  we  obtain  Sj  =  —  2,  S2  =  6,  S3  =  —  31/2. 

Therefore,  §814,  Sl/a  =  -2,  Sl/a2  =  6,  Sl/a3  =  _3i/2. 


THEORY   OF    EQUATIONS  481 

Example  2.  For  the  equation  /(i)  =  x*  +  6ix^  +  h2x'^  +  63X  +  64  =  0, 
show  that 

Si  +  4  61  =  0  6i,  S2  +  fti'Si  +  4  60  =  2  62,    Ss  +  61S2  +  62S1  +  4  63  =  63, 
Si  +  61S3  +  62S2  +  63S1  +  4  64  =  0,  85  +  61S4  +  62S3  +  63S2  +  64S1  =  0, 

and  compute  Sj,  S2,  S3,  S4  in  terms  of  61,  625  ^3,  ^4- 

The  preceding  formulas  also  show  that  s^,  s^,  S3,  •  •  ■  are  infe-     867 
f/ral  functions  of  the  coefficients  of  the  equation  when,  as  in 
(1),  the  leading  coefficient  is  1. 

Theorem  2.     Every  rational  symmetric  function  of  the  roots     868 
of  an  equation  f(x)=0  can  be  expressed  rationally  in  terms 
of  its  coefficients. 

Let  the  roots  of  f(x)  =  0  be  or,  ;8,  y,  •  •  • ,  v. 

Every  rational  symmetric  function  of  a,  /3,  •••,  v  can  be 
expressed  rationally  in  terms  of  functions  of  the  types  Sor'', 
Sa"/??,  '^crJ' fi'J-f,  and  so  on,  §  544.  Hence  it  is  only  necessary 
to  prove  our  theorem  for  functions  of  these  several  types. 
This  was  done  for  the  type  2tt"  =  s^  in  §  866,  and  we  shall 
now  show  that  2«''^,  and  so  on,  can  be  expressed  rationally 
in  terms  of  functions  of  this  type  s^,. 

1.    The  type  2a'^/ff'  =  a''/?'  +  /J^Vt'  -\ . 

The  product    (a'^  +  yS''  H )  (a'  +  /J'  H )  (1) 

is  the  sum  of  the  two  symmetric  groups  of  terms 

aP  +  n^pp  +  1^ ,  (2) 

N  a^P^  +  I3"a''  +  ---.  (3) 

But  (1)  and  (2)  are  rational  functions  of  the  coefficients, 
§  866.  Hence  (3),  or  Sa''/?',  which  may  be  obtained  by  sub- 
tracting (2)  from  (1),  is  also  such  a  function. 

Since  (1)  is  s^s^  and  (2)  is  s^^^,  we  have  the  formula 

^"''^^V.-^Pfr  (4) 

When  p  =  q,  the  terms  of  (3)  are  equal  in  pairs  and  (3) 
becomes  2  Sa^/S^.     We  then  have  2  %aPpp  =  sj  —  s^p^ 


482  A  collp:ge  algebra 

2.  The  type  Sa'^yS'^y'"  =  a'^l^y  +  /?'aY  H • 

The  product  (a''/?'  +  ^''a''  -] )  (a--  +  /?'-  +  y'-  h )  (5) 

is  the  sum  of  the  three  symmetric  groups  of  terms 

^7.+  r^7^^,.  +  r,^7_j ^  (6) 

a^'^'  +  '-  +  y8"a'  +  '-H ,  (7) 

«'')8Y"  +  /3''aV +••••  (8) 

But  we  have  already  shown  that  (5),  (6),  and  (7)  are  rational 
functions  of  the  coefficients.  Hence  (8),  or  Sa^yS^y'",  which 
may  be  obtained  by  subtracting  the  sum  of  (6)  and  (7)  from 
(5),  is  also  such  a  function. 

When 2^  =  q,  the  group  (8)  becomes  2 Sa^'yS^y'';  when p  =  q=zr, 
it  becomes  6  Sa'^^S^'y''. 

3.  The  types  ^a''(3Y^",  and  so  on. 

We  may  prove  that  these  are  rational  functions  of  the  coeffi- 
cients by  repetitions  of  the  process  illustrated  in  1  and  2.  We 
begin  by  multiplying  2a''j8''y''  by  a"  +  /3"  +  y"  •  •  • . 

Example.     Show  that 

I,aP^iy  =  Sj,SqSr  +  2Sj,  +  ,i  +  r  -  Sj,  +  rjSr  -  S^  +  rSp  -  Sr  +  pS,. 

EXERCISE  LXXIX 

1.  For  the  equation  aoz^  +  aiX^  +  a2X  +  as  =  0  find  S3  and  84  in  terms 

of  ao,  «1,  «2i   <^3- 

2.  If  a,  p,  7  denote  the  roots  of  x^  +  px2  +  qx  +  r  -  0,  find  S  1/a^, 
2  1/  a^,  and  S  afi'^  in  terms  of  p,  q,  r. 

3.  Find  the  equation  whose  roots  are  the  cubes  of  the  roots  of 
a;3_2x2  +  3x-l  =  0. 

4.  If  a,  13,  7  denote  the  roots  of  the  equation  x^  -  x^  +  3  x  +  4  =  0, 
find  the  values  of  the  following  symmetric  functions  of  these  roots  by  the 
methods  of  §§  866,  867. 

(1)  si,  So,  83,  84.  (2)  Sa3/J2.  (3)  Sa3^7. 

(4)  Sa3i327.  (5)  21/ a*.  (6)  Sa^^/V. 


CUBIC    AXD   BIQUADRATIC    EQUATIONS  483 

XXX.     THE    GENERAL    CUBIC    AND 
BIQUADRATIC    EQUATIONS 

Algebraic  solutions.  In  the  preceding  chapter  we  have  shown  869 
that  the  real  roots  of  a  numerical  equation  can  always  be 
found  exactly  or  approximately,  and  it  is  possible  to  extend 
the  methods  there  employed  to  the  complex  roots  of  such  equa- 
tions. As  hardly  need  be  said,  these  methods  are  not  appli- 
cable to  literal  equations.  To  solve  such  an  equation  we  must 
obtain  expressions  for  its  roots  in  terms  of  its  coefficients. 

We  say  that  an  equation  can  be  solved  algebraically  when 
its  roots  can  be  expressed  in  terms  of  its  coefficients  by  apply- 
ing a  finite  number  of  times  the  several  algebraic  operations, 
namely,  addition,  subtraction,  multiplication,  division,  involu- 
tion and  evolution. 

We  have  already  proved,  §  631,  that  the  general  quadratic 
equation  has  such  an  algebraic  solution,  and  we  are  now  to 
prove  that  the  like  is  true  of  the  general  cubic  and  biquadratic 
equations.  But  general  equations  of  a  degree  higher  than  four 
cannot  be  solved  algebraically. 

Cube  roots  of  unity.  In  §  Q^Q  we  showed  that  the  equa-  870 
tion  x^  =  1  has  the  roots  !,(-!+  i  V3)/2,  (-  1  -  i  V3)/2. 
Hence  each  of  these  numbers  is  a  cube  root  of  unity.  The  third 
will  be  found  to  be  the  square  of  the  second.  Hence  if  we 
represent  the  second  by  w,  we  may  represent  the  third  by  w^. 
Since  x^  —  1  =-0  lacks  an  x^  term,  we  have,  §  805, 1  +  w  +  w^  =  0. 

Similarly  every  number  a  has  three  cube  roots,  namely,  the 
three  roots  of  the  equation  x^  —  a.  If  one  of  these  roots  be 
Va,  the  other  two  will  be  w  Va  and  w^  Va. 

The  general  cubic.    Cardan's  formula.     By  the  method  of  §  818     871 
every  cubic  equation  can  be  reduced  to  the  form 

x^+px  +  q  =  0,  (1) 

in  which  the  x"^  term  is  lacking. 


484  A   COLLEGE    ALGEBRA 

In  (1)  put  x=^y  +  z.  (2) 

We  obtain 

tf  +  ^,fz^Syz^  +  z'^+p{y  +  z)+q  =  0, 
or  y^  +  z^+(Zyz+p){y^z)  +  q  =  Q.  (3) 

As  the  variables  y  and  z  are  subject  to  the  single  condition 
(3),  we  may  impose  a  second  condition  upon  them. 

We  suppose  ^yz  +  p  =  0,  (4) 

and  therefore,  by  (3),  y^  +  z""  +  q  =  0.  (5) 

From  (5),  y^  +  z''  =  -q,  (6) 

and  from  (4),  y^z^=-p^l21.  (7) 

Therefore,  §  636,  y^  and  z^  are  the  roots  of  a  quadratic  equa- 
tion of  the  form 

,,2  +  ^„_^3y27  =  0.  (8) 

Solving  (8)  and  representing  the  expressions  obtained  for 
the  roots  by  A  and  B  respectively,  we  have 


2/' 


These  equations  (9)  give  three  values  for  y  and  three  for  s, 
namely,  §  870, 

y  =  ^,    wVCT,  0)2^,  (10) 

z  =  ^,    w-^B,    u)2^.  (11) 

But  by  (4),  yz  =  —  p/3,  and  the  only  pairs  of  the  values 
of  y,  z  in  (10),  (11)  which  satisfy  this  condition  are 

y,  z  =  -y/A,   -y/B  ;  w  -y/l,  w-  -^/b  ;  w^  "vCT,  w  ^/b. 

Substituting  these  pairs  in  (2),  we  obtain  the  three  roots  of 
(1),  namely, 

iCi  =  V^  +  -y/Ji,  Xo  =  w  V^  J  +  to^  ^B,  Xs  =  <o2  vCT  +  a>  -^B, 
where      .,  =  _  |  +  ^fTI.  B  =- |  -  Vf^l-  (12) 


CUBIC    AND    BIQUADRATIC    EQUATIONS  485 

Example  1 .     Solve  x^  -  6  x^  +  6  x  -  2  =  0. 

By  §  818  we  find  that  the  substitution  x  =  y  -\-  2  will  transform  the 
given  equation  to  the  form  y^  +  py  +  q  =  0.     We  thus  obtain 
2/3  _  G  ?/  -  6  =  0. 

The  roots  of  this  equation,  found  by  substituting  p=—  6,  g  =  — 6in 
the  above  formulas,  are 

3,—  3,—       3/—  a  r-  ^r-  '^  r- 

V4  +  V2,    V4w  +  V2w2,    V4a;2  +  V2w. 
Hence  the  roots -of  the  given  equation  are 

2  +  v'i  +  V2,  2  •+ V^  t^  +  V2  u)2,  2  +  \/4  a.2  + 'v'i  w. 
Example  2.     Solve  the  equation  x^  -\-  o  x"^  -\-  Q  x  +  b  =  0. 

Discussion  of  the  solution.     When  p  and  q  are  real,  the  char-     872 

acter  of  the  roots  depends  on  tlie  value  of  y^/4  +^j^/27  as 
follows  : 

1.  i,"^  q^/4  +  p^/27  >  0,  one  root  is  real,  two  are  imaginary. 

For  in  this  case  A  and  B  are  real.  Hence  x^  is  real,  and  x^, 
^3  are  conjugate  imaginaries,  §  870. 

2.  If  q^/4  -f-  p'/27  =  0,  all  the  roots  are  real  and  two  equal. 
For  in  this  case  A  =  B  =  —  q /2.     Hence  rri=— 2V^/2, 

3/ 3/ 

and  X.2  =  iCj  =  —  (w  +  or)  V y/2  =  \q/2,  since  w  +  w'^  =  —  1, 
by  §  870. 

3.  i/'q"/4  +  p^/27  <  0,  all  the  roots  are  real  and  unequal. 

This  may  be  proved  by  Sturm's  theorem  (see  p.  477,  Ex.  16). 
I>ut  when  q^ / ^  +p^ /21  <  0,  A  and  B  are  complex  numbers, 
and  though  the  expressions  for  a-j,  X2,  Xs  denote  real  numbers, 
they  cannot  be  reduced  to  a  real  form  by  algebraic  transforma- 
tions. This  is  therefore  called  the  irreducible  case  of  the 
culuc  (see  §  885). 

The  expression  q^/i  +^j^/27  is  called  the  discriminant  of     873 
the  cubic  x^  +/?x  -f  5-  =  0,  since  its  vanishing  is  the  condition 
that  two  of  the  roots  be  equal  (compare  §  635). 


486  A   COLLEGE    ALGEBRA 

874         The  general  biquadratic.    Ferrari's  solution.     By  the  method  of 
§  818  every  biquadratic  equation  can  be  reduced  to  the  form 

;  X*  +  ax^  +  bx  +  c  =  0.  (1) 

With  a  view  to  transforming  the  first  member  of  (1)  into  a 
difference  of  squares,  we  add  and  subtract  xhi  +  i<^/4,  where 
u  denotes  a  constant  whose  value  is  to  be  found.  We  thus 
obtain 

_  X*  +  x^-u  +  u^/i  -  xhi  —  ?<V4  +  ax'^  +  hx  +  c=^0, 

or  (x"  +  u/2y  -  [(m  -  a)  x""  -  bx  +  (i*V4  -  c)]  =  0.      (2) 

To  make  the  second  term  a  perfect  square,  we  must  have 

b-  =  ^{u-a){u^/4.  -c), 

or  u^  -  ait"  -4.CU+  (4  ac  -  b"")  =  0.  (3) 

Let  «i  denote  one  of  the  roots  of  this  cubic  in  «.  When  u 
is  replaced  by  ?/i  in  (2),  the  second  term  is  the  square  of 
V?ii  —  ax  —  b /2  V??i  —  a,  and  (2)  becomes  equivalent  to  the 
two  quadratics 

^^  +  v;;;^:^  ^  +  (t  -  ^^~ — ^  =  0.         (4) 

'"i    .  b 

2  Vi^i  -  a^ 

We  may  therefore  obtain  the  roots  of  (1)  by  solving  (4) 
and  (5). 

Example  1.     Solve  x*  +  x"^  +  4x  -  S  =  0. 

Here  a  =  1,  6  =  4,  c  =  —  3,  so  that  the  cubic  (3)  is 

M«  -  rt2  +  12  u  -  28  =  0. 
One  of  the  roots  of  this  cubic  is  2,  and  setting  wi  =  2  in  (4),  (5),  we  have 

^2  +  x  -  1  =  0  and  x^  -  x  +  3  =  0. 
Solving  these  quadratics,  we  obtain  x  =  (-  1  ±  V5)/2,  (1  -t  i  \^)/2. 
Example  2.     Solve  the  equation  x*  —  4  x''  +  x^  +  4  x  +  1  =  0. 

As  the  cubic  (3)  has  three  roots  any  one  of  which  may  be 
taken  as  the  Ui  in  (4)  and  (5),  it  would  seem  that  the  method 


V^^r^a:+f^  +  _/ )^0.  (5) 


CUBIC    AND    BIQUADRATIC    EQUATIONS  487 

above  described  might  3'ield  3-4  or  12  values  of  x,  whereas 
the  given  equation  (1)  can  have  but  four  roots.  But  it  is  not 
difficult  to  prove  that  the  choice  made  of  u^  does  not  affect 
the  values  of  the  four  roots  of  (4),  (5)  combined,  but  merely 
the  manner  in  which  these  roots  are  distributed  between  (4) 
and  (5). 

Reciprocal  equations.  We  may  discover  by  inspection  whether  87fi 
or  not  a  given  reciprocal  equation,  §  815,  has  either  of  the  roots 
1  or  —  1,  and  if  it  has,  derive  from  it  a  depressed  equation 
</)  (.'•)  =  0  which  has  neither  of  these  roots.  It  follows  from 
§815  that  this  depressed  equation  ^  (x)  =  0  must  be  of  the 
form 

ciqX^"'  +  aia;-'"- '  +  •  ■  •  +  «'„,^'"'  +  ■  •  ■  +  a^x  +  a^^  —  0,      (1) 

that  is,  its  degree  must  be  even,  and  every  two  coefficients 
which  are  equally  removed  from  the  beginning  and  end  of 
(^(x)  must  be  equal. 

\\%  proceed  to  show  that  by  the  substitution  z  =  x  -{-\  jx 
this  equation  ^(a-)=  0  may  be  transformed  into  an  equation 
in  z  whose  degree  is  one  half  that  of  <f)(x)=  0.  It  will  then 
follow  that  if  the  degree  of  <j>(x)=  0  be  not  greater  than  eight, 
we  may  find  the  roots  by  aid  of  §§  631,  871,  874. 

Dividing  (1)  by  x"'  and  combining  terms,  we  have 

But  by  carrying  out  the  indicated  reckoning  we  find  that 

and  if  z  =  x  +  1/x,  and  in  (3)  we  set  jc)  =  1,  2,  3  •  •  •  succes- 
sively, we  have 

x'  +  ~  =  z'-  2,  x^  +  \^z^-Sz, 
x^  x^ 

x'  +  ^,  =  z^-^z^  +  2,>..,  (4) 


488 


A   COLLEGE   ALGEBRA 


876 


877 


that  is,  we  obtain  for  a^P  +  l/x"  an  expression  of  the  j9th 
degree  in  z.  Substituting  these  several  expressions  in  (2),  we 
obtain  an  equation  of  the  mth  degree  in  »,  as  was  to  be  demon- 
strated (compare  §  645). 

Example  1.     Solve  2x8  _  x7  _  12  x6  +  Hx^  -  14a;3  +  12x2  +  x  -  2  =  0, 
This  is  a  reciprocal  equation  having  the  roots  1  and  —  1. 
Removing  the  factor  x^ 


2x6 


1, 
lOx^  +  13x3  -  10x2  -  X  +  2  =  0. 


Dividing  by  x^, 


13  =  0. 


Hence  by  (4),         2  (z^  -  3  2)  -  (f  -  2)  -  10  z  +  13  =  0, 
or  2z3-22_i6z  +  15  =  0. 

Solving,  z  =  1,   -3,  or  5/2. 

Hence  x  +  l/x  =  l,   -3,  or  5/2, 

andtherefore  x=(l  ±i  V3)/2,   (-3±V5)/2,  2,  or  1/2 

Example  2.     Solve  x^  -  x^  +  x*  +  x2  -  x  +  1  =  0. 

Every  binomial  equation  «"  +  a  =  0  may  be  reduced  to  the 
reciprocal  form  by  aid  of  the  substitution  x 


of  this  substitution  being  y"  +  1 


Vay,  the  result 
0  (compare  §  646,  Ex.  2). 


Expression  of  a  complex  number  in  terms  of  absolute  value  and 
amplitude.     In  the  accompanying  figure,  P  is  the  graph  of  the 


^ 

P 

li 

/ 

h 

/r 

1 

A 

0 

a      ^ 

complex  number 
in  §  238. 


+  bi,  constructed 


Tlie  length  of  OP  is  Vci^  +  b%  the 
absolute  value  of  a  +  bi,  §  239.  Repre- 
sent it  by  r. 

Let  6  denote  the  circular  measure  of 
the  angle  XOP,  that  is,  the  length  of  the 
arc  subtended  by  this  angle  on  a  circle  of 
unit  radius  described  about  0  as  center. 
We  call  6  the  amplitude  of  a  +  bi. 

We  call  the  ratio  b/r  the  sine  of  0,  and  write  b/r  =  sin  0. 


CUBIC    AND    BIQUADRATIC    EQUATIONS  489 

We  call  the  ratio  a/r  the  cosine  of  6,  and  write  a  fr  =  cos^. 

We  thus  have  a  —  r  cos  6,  h  —  r  sin  0, 

and  therefore         a  -\-hi  ^=  r  (cos  6  +  i  sin  6). 

When  ^  =  0,  then  b  =  0  and  a  =  r.     Hence  sin  0  =  0,  and 
cosO  =  1. 

The  circular  measure  of  360°  is  2  tt,  this  being  the  length     878 
of  a  circle  of  unit  radius.     Hence  a  point  P  given  by  r,  6  is 
given  equally  by  r,  ^  +  2  tt  ;  by  ?•,  ^  —  2  tt  ;  and,  in  general,  by 
r,  0  -{-  2  niTT,  where  m  denotes  any  integer.     Hence  we  say  that 
the  general  value  of  the  amplitude  of  a  +  ^*  is  ^  +  2  tnir. 

Theorem.     The  absolute  value  of  the  product  of  two  complex     879 
numbers  is  the  product  of  their  absolute  values;  and  its  ampli- 
tude is  the  sum  of  their  amplitudes. 

For  r  (cos  6  -\-  i  sin  6)  ■  r'  (cos  &  +  /  sin  6') 

=  r  /•'  [(cos  6  cos  6'  —  sin  0  sin  6') 

+  i  (sin  e  cos  6'  +  cos  0  sin  $')'] 
=  r  r'  [cos  (0  +  0')  +  i  sin  {d  +  6')], 

since  it  is  proved  in  trigonometry  that 

cos  (6  +  6')  =  cos  6  cos  0'  -  sin  $  sin  6', 
sin  (6  +  6')  =  sin  0  cos  0'  +  cos  0  sin  ^'. 

The  construction  given  in  §  240  is  based  on  this  theorem. 
Corollary  1.     By  repeated  applications  of  §  879  we  have  880 

r  (cos  6  +  i  sin  6)  ■  r'  (cos  6'  +  i  sin  0')  ■  r"  (cos  0"  +  i  sin  ^")  •  •  • 
=  rr'/-".--[cos(^  +  ^'  +  ^"  +  ...) 

+  i  sin  (e  +  O'  +  6"  +  ■•  •)]. 

Corollary  2.     Setting  r  =  r' =  r"  =  •••,  and  ^  =  ^' =  ^"  =  ••  •     881 

in  §  880,  we  obtain  the  following  formula,  known  as  Demoivre's 
theorem  : 

[?'(cos  6  +  I  sin  ^)]"  =  ?•"  (cos  nO  +  i  sin  nO). 


490  A   COLLEGE   ALGEBRA 

Corollary  3.     For  a  quotient  we  have  the  formula 

r  (cos  ^  +  i  sin  ^)         i-  ,^       ^,,       .    ■     ,^       ^„-, 

-^ .,  ;    .    ■     .;  =  -  [cos  {Q  -  6')  +  i  sm  {d  -  6')\ 

r' (cos  ^' +  i  sm  ^')       ?•"-       ^  ■'  \  jj 

For  ^  [cos  {6  -  $')  +  i  sin  {6  ~  6')']-  r'  (cos  6'  +  i  sin  0') 

^r(cose  +  isme).  §879 

Corollary  4.    Tlie  n  /;th  roots  of  a  complex  nvmiberare  given  by 

n/     ,  .      ■       ,  V  ^  +  2  A-TT 

v  ?•  (cos  ^  +  i  sni  ^j  =  y"  I  cos 

when  k  is  assigned  the  ?i  values  0,  1,  2,  • 

rW       e  +  2k7r   ,    .    .    ^  +  2A:7r\"|" 
For  ?•"    cos —  +  i  sm 

=  r  [cos  (^  +  2  A-tt)  +  i  (sin  ^  +  2  Att)]  =  r  (cos  O^i  sin  ^). 

§§  881,  878 

Binomial  equations.     The  n  nth.  roots  of  r  (cos  6  +  i  sin  6)  are 

the  n  roots  of  the  equation  x"  —  ?'(cos  ^  +  t  sin  ^)  =  0.     Hence, 

in  particular,  the  n  roots  of  the  equation  x"  —  r  =  0,  where  r  is 

real  and  therefore  6  is  0,  are 

V^(cos  2  kiT In  +  i  sin  2  kir  fn),  k  =  0,  1,  •  •  •,  n  —  1. 

Thus,  the  roots  of  the  equation  x'  —  1  =  0  are 

cosO+isinO,  cos2  tt/S  +  i  sin  2  7r/3,  cos4  7f/3  +  isin4  tt/S, 
which  may  be  proved  equal  to 

1,     (-l  +  zV3)/2,     (-l-iV3)/2. 
Trigonometric  solution  of  the  irreducible  case  of  the  cubic.     In 
the  irreducible  case  of  the  cubic  x^  +  px  -{-  q  =  0,  §  872,  3,  the 
expressions  A  and  B,  §  871,  are  conjugate  imaginaries.     For 
since  in  this  case  rf/i  +2^^/27  is  negative,  we  have 


^=-2+,- 


■■V-(^0— i-W-e-^ 


Hence  the  expressions  for  A  and  B  in  terms  of  absolute 
value  and  amplitude,  §  877,  will  be  of  the  form 

^  =  r  (cos  e  +  i  sin  6),  B  ^  r(cos  6  -i  sin  6),  (1) 


CUBIC    AND    BIQUADRATIC    EQUATIONS  491 

where 


—  q  rl 
and  cos  6  =  — — — 


ff-MyK-0" 
"km 


(2) 


When  J)  and  q  are  given  we  can  find  the  value  of  Q  from 
tliat  of  cos  Q  by  aid  of  a  table  of  cosines. 

In  the  formulas  for  the  roots   of  x^  -^  px  -\-  q  =  ^,  §  871, 

(12),  substitute   the    expressions  (1)  for  A    and  i?,  and  the 

expressions  for  w  and  w'-^  given  in  §  884.     The  results  when 

simplified  are 

o   t        ^              o   i        ^  +  27r              ^   .        ^  +  47r 
jCi  =  2  /•'  cos  - )  a:'o  =  2  r^  cos >  cr,  =  2  r^  cos 

And,  ?•  and  Q  being  known  by  (2),  these  formulas  enable  us 
to  compute  the  values  of  the  roots  by  aid  of  a  table  of  cosines. 

Example.     Solve  x3-x  +  l/3  =  0. 

Here  Q'2/4  +  -p^ /11  =  —  1/108,  so  that  we  have  the  irreducible  case. 

Substituting  in  the  formulas  (2)  and  simplifying,  we  find 

r-\/  V27,  cos  6»  =  -  V3/2,  and  therefore  B  =  150°. 
Hence  by  aid  of  tables  of  logarithms  and  cosines  we  obtain 

Xx  =  -^  cos  50°  =  .  7422  ;  x^  =  -^  cos  170°  =  -  1. 1371 ; 

V27  V27 

X3  =  ^  cos  290°  =  .3949. 

V27 

EXERCISE   LXXX 

Solve  equations  1  —  10  by  the  methods  of  §§  8T1  and  874. 

1.   x3-9x-28  =  0.  2.    x3-9x2  +  9x-8  =  0. 

3.    x3  -  3  X  -  4  =  0.  4.    4  a;3  -  7  X  -  6  =  0. 

5.    x3  +  3x2  +  9x-l  =  0.  6.    3x5-0x2  + 14x +  7  =  0. 

7.    X*  +  x2  +  6  X  +  1  =  0.  8.    x^  -  4  x3  +  x2  +  4  X  +  1  =  0. 

9.    x*  +  12x-5  =  0.  10.   x<  +  8x3  +  12x2-l)x  +  2  =  0. 

11.    Solve  3x6-2x6  +  6x*-2x»  +  6x2-2x  +  3  =  0, 


492  A   COLLEGE    ALGEBRA 

12.  Solve  2x8  -  9x7+18x6  _  sOx^  +  32x4  -  30x^  +  18x2  -  9x  +  2  =  0. 

13.  Solve  6  x"  -  x6  +  2  x5  -  7  X*  -  7  x3  +  2  x2  -  X  +  6  =  0. 

14.  Find  the  cubic  in  z  on  which,  by  §  875,  the  solution  of  x^  -  1  =  0 
depends. 

15.  Find  the  condition  that  all  the  roots  of  x^  +  3  ax^  +  3  to  -r  c  =--  0 
be  real. 

16.  Write  down  the  trigonometric  expressions  for  the  roots  of  x''  -  1=0, 
and  of  x6  +  1  z=  0. 

17.  Solve  the  following  irreducible  cubics. 

(1)  x3  -  3 X  -  1  =  0.  (2)  x3  -  6x  -  4  =  0. 

18.  In  a  sphere  whose  diameter  is  3  %/3  a  right  prism  with  a  square 
base  is  inscribed.     If  the  volume  of  the  prism  is  27,  what  is  its  altitude  ? 

19.  The  volume  of  a  certain  right  circular  cylinder  is  50  7t  and  its 
entire  superficial  area  is  105  tt/ 2.  Find  the  radius  of  its  base  and  its 
altitude. 

20.  The  altitude  of  a  right  circular  cone  is  6  and  the  radius  of  its  base 
is  4.  In  this  cone  a  right  circular  cylinder  is  inscribed  whose  volume  is 
four  ninths  that  of  the  cone.     Find  the  altitude  of  the  cylinder. 


XXXI.     DETERMINANTS    AND   ELIMINATION 

DEFINITION   OF   DETERMINANT 

886  Inversions.  Odd  and  even  permutations.  When  considering 
the  permutations  of  a  set  of  objects,  as  letters  or  numbers, 
we  may  fix  upon  some  particular  order  of  the  objects  as  the 
normal  order.  Any  given  permutation  is  then  said  to  have  as 
many  inversions  as  it  presents  instances  in  which  an  object  is 
followed  by  one  which  in  the  normal  order  precedes  it.  And 
the  permutation  is  called  odd  or  even  according  as  the  number 
of  its  inversions  is  odd  or  even  (or  0). 

Thus,  if  the  objects  in  normal  order  are  the  numbers  1,  2,  3,  4,  5,  the 
permutation  45^12  lias  the  eight  inversions  43,  41,  42,  53,  51,  52,  31,  32, 
Hence  45312  is  an  even  permutation. 


DETERMINANTS    AND   ELIMINATION  493 

Theorem,     If  two  of  the  objects  in  a  permutation  are  inter-     887 
changed,  the  number  of  inversions  is  increased  or  diminished  by 
an  odd  number. 

For  if  two  adjacent  objects  are  interchanged,  the  number  of 
inversions  is  increased  or  diminished  by  1.  Thus,  compare 
ApqB  (1)  and  AqpB  (2),  where  A  and  B  denote  the  groups  of 
objects  which  precede  and  follow  the  interchanged  objects  p 
and  q.  Any  inversions  which  may  occur  in  A  and  B  and  any 
which  may  be  due  to  the  fact  that  A,  p  and  q  precede  B  are 
common  to  (1)  and  (2).  .  Hence  the  sole  difference  between 
(1)  and  (2),  so  far  as  inversions  are  concerned,  is  this  :  li  pq  is 
an  inversion,  qp  is  not,  and  (2)  has  one  less  inversion  than  (1) ; 
but  if  pq  is  not  an  inversion,  qj)  is,  and  (2)  has  one  more 
inversion  than  (1). 

But  the  interchange  of  any  two  objects  may  be  brought 
about  by  an  odd  number  of  interchanges  of  adjacent  objects. 
Thus,  from  pabq  we  may  derive  qahp  by  five  interchanges  of 
adjacent  letters.  We  first  interchange  p  with  each  following 
letter  in  turn,  obtaining  successively  aphq,  abpq,  abqp,  and  we 
then  interchange  q  with  each  preceding  letter,  obtaining  aqhp, 
qabp.  There  is  one  less  step  in  the  second  part  of  the  process 
than  in  the  first  part,  because  when  it  is  begun  q  has  already 
been  shifted  one  place  in  the  required  direction.  Had  there 
been  /x  letters  between  p  and  q,  there  would  have  been  /a  +  1 
steps  in  the  first  part  of  the  process  and  /a  in  the  second,  and 
(/A  +  1)  +  ^,  or  2  /A  +  1,  is  always  odd. 

Therefore,  since  each  interchange  of  adjacent  objects  changes 
the  number  of  inversions  by  1  or  —  1,  and  the  sum  of  an  odd 
number  of  numbers  each  of  which  is  1  or  —  1  is  odd,  our 
theorem  is  demonstrated. 

Thus,  if  in  21457368  (1)  we  interchange  4  and  6,  we  get  21657348  (2). 
It  will  be  found  that  (1)  has^we  inversions  and  (2)  eight,  and  8  —  5  is  odd. 

Of  the  n\  permutations  of  n  objects  taken  all  at  a  time,     888 
§  763,  half  are  odd  and  half  are  even.     For  from  any  one  of 


494  A   COLLEGE   ALGEBRA 

these  permutations  we  can  derive  all  the  rest  by  repeated 
interchanges  of  two  objects.  As  thus  obtained,  the  permuta- 
tions will  be  alternately  odd  and  even,  or  vice  versa,  §  887. 
Therefore,  since  n !  is  an  even  number,  half  of  the  permutations 
are  odd  and  half  are  even. 

889  In  what  follows  we  shall  have  to  do  with  sets  of  letters 
with  subscripts,  as  a^,  a.^,  ■■-,  h^,  bo,  ■■■,  and  so  on.  Having 
chosen  any  set  of  such  symbols  in  which  all  the  letters  and 
subscripts  are  different,  arrange  them  in  some  particular  order 
and  then  find  the  sum  of  the  number  of  inversions  of  the 
letters  and  of  inversions  of  the  subscripts.  If  this  sum  is 
even,  it  will  be  even  when  the  symbols  are  arranged  in  any 
other  order ;  if  odd,  odd.  For  when  any  two  of  the  symbols 
are  interchanged  the  inversions  of  both  letters  and  subscripts 
are  changed  by  odd  numbers,  §  887,  and  therefore  their  sum 
by  an  even  number. 

In  particular,  the  number  of  inversions  of  the  subscripts 
when  the  letters  are  in  normal  order  and  that  of  the  letters 
when  the  subscripts  are  in  normal  order  are  both  odd  or  both 
even. 

Thus,  in  a-^bsCi  the  number  of  inversions  of  the  subscripts  Is  two; 
in  Cia2&3  the  number  of  inversions  of  the  letters  is  tioo ;  in  ftsOoCi  the 
sum  of  the  number  of  inversions  of  the  letters  and  that  of  the  subscripts 
is  four. 

890  Definition  of  determinant.     We  may  arrange  any  set  of  2^,  3^ 

in  general  n'^  numbers  in  the  form  of  a  square  array,  thus : 


fll     ^2 

(h 

(Xo  a  3 

«i 

«2     fiz    ^4 

h,     b. 

h 

h   b. 

h 

h^     bs     64 

Cl 

Co      C3 

^2     Cs     Ci 
d.     d,    d. 

and  so  on,  where  the  letter  indicates  the  roiv,  and  the  subscript 
the  column,  in  which  any  particular  number  occurs. 

We  may  then  call  the  numbers  the  elements  of  the  array. 


DETERMINANTS    AND    ELIMINATION  495 

With  the  elements  of  such  an  array  form  all  the  jJroducts  that 
can  he  formed  by  taking  as  factors  one  element  and  but  one  from 
each  roiv  and  each  column  of  the  array. 

In  each  product  arrange  the  factors  so  that  the  row  marks 
(letters^  are  in  normal  order,  and  then  count  the  inversions  of 
the  column  marks  (subscripts).  If  their  number  be  even  (or  0), 
give  the  product  the  plus  sigri;  if  odd,  the  minus  sign. 

The  algebraic  sum,  of  all  these  plus  and  minus  jjroducts  is 
called  the  deteiininant  of  the  array,  and  is  represented  by  the 
array  itself  with  bars  written  at  either  side  of  it. 

Thus,  L  ^   ,  ^    =  «1&2  —  «2^1 


and 


«!  a-i  as 
61  62  ^3 

Ci    C2    Cz 


61  62 
aih^Cz  -  aibzC2  +  a^hiCi  —  aobiCz  +  a3biC2  —  (136201. 


It  follows  from  §  889  that  in  determining  the  sign  of  any     891 
of  the  products  above  described  we  may  arrange  the  factors  so 
that  the  column  marks  are  in  normal  order,  and  then  give  the 
product  the  plus  or  minus  sign  according  as  the  number  of 
inversions  of  the  row  marks  is  even  or  odd. 

Or  we  may  write  the  factors  in  any  order  whatsoever,  and 
then  give  the  product  the  plus  or  minus  sign  according  as  the 
su7)i  of  the  number  of  inversions  of  row  marks  and  of  column 
marks  is  even  or  odd. 

Thus,  consider  the  product  Uzb^Ci  to  which  our  first  rule  gave  the  minus 
sign.  Writing  it  so  that  tlie  subscripts  are  in  normal  order,  we  have 
Cib2as,  and  since  the  letters  cba  now  present  three  inversions,  the  product 
must  again,  by  our  second  rule,  be  given  the  minus  sign.  If  the  prod- 
uct be  written  hoCias,  the  letters  present  two  inversions  and  the  subscripts 
one,  and  2  +  1  is  odd.  Hence  again,  by  the  third  rule,  we  must  give  the 
product  the  minus  sign. 

The  number  of  the  rows  or  columns   in  the  array  out  of     892 
which   a  determinant   is   formed   is   called   the  oi-der  of  the 
determinant. 


496  A    COLLEGE    ALGEBRA 

893  The  products  above  described,  with  their  proper  signs,  are 
called  the  terms  of  the  determinant. 

894  To  expand  a  determinant  is  to  write  out  its  terms  at 
length. 

895  The  diagonal  of  elements  a^,  Z»2,  c^,  ■■■  is  called  the  leading 
diago7ial,  and  the  product  a-Jj^c^,  ■  •  •  is  called  the  leading  term 
of  the  determinant. 

The  leading  term  enclosed  by  bars,  thus,  \a^  h^  ^3  •  •  -I,  is  often 
used  as  a  symbol  for  the  determinant  itself. 

896  The  number  of  the  terms  of  a  determinant  of  the  nth  oi-der 
is  n !,  and  half  of  these  terms  have  plus  signs  and  half  have 
minus  signs. 

Por,  keeping  the  letters  in  normal  order,  we  may  form  n\ 
permutations  of  the  n  subscripts,  §  763,  and  there  is  one 
term  of  the  determinant  for  each  of  these  n\  permutations, 
§890. 

Furthermore  half  of  these  n\  permutations  are  even  and 
half  are  odd,  §  888. 

Thus,  for  n  =  3  we  have  3  !  or  6  terms ;  for  n  =  4,  we  have  4  !  or  24. 

897  Other  notations.  It  must  be  remembered  that  the  letters  and 
subscripts  are  mere  marks  of  row  and  column  order.  Any 
other  symbols  which  will  serve  this  purpose  may  be  substi- 
tuted for  them. 

Thus,  the  elements  of  a  determinant  are  often  repre- 
sented by  a  single  letter  with  two  subscripts,  as  023,  the 
first  indicating  the  row  and  the  second  the  column.     ThR 
symbol  a^z  is  read  "  a  two  three,"  and  so  on. 

898  A  rule  for  expanding  a  determinant  of  the  third  order.  To  obtain 
tlie  three  positive  terms,  start  at  each  element  of  the  first  row 
I  «i  «2  03     ^"^  *^^^  ^^^'  ^°  ^^^  ^^  possible,  follow  the  direction 

i>i  ^2   ^-'3      ^^  ^^^®  leading  diagonal,  thus  : 
ri   ^2   C3  a^b^Ci,  azb^Ci,  a^biC^. 


an 

«12 

ai3 

ail 

a-n 

«23 

asi 

az2 

033 

DETERMIXANTS    AND   ELIMIXATION 


497 


To  obtain  the  three  negative  terms,  proceed  in  a  similar 
manner  but  follow  the  direction  of  the  other  diagonal,  thus : 

—  «1^3f'2,     —   «2^'l^3,     —  «3^2^1- 


5       3 

1  - 1 

2  4 


5(-l)(-l)  +  3(-3)2  +  2(-l)4 

-5(-3)4-3(-l)(-l)-2(-l)2  =  40. 


This  rule  does  not  apply  to  determinants  of  an  order  higher  than  the 
third.  Thus,  it  would  give  but  eight  of  the  twenty-four  terms  of  a  deter- 
minant of  the  fourth  order. 


EXERCISE  LXXXI 

Expand  the  following  determinants. 


p       q        r 

1         X         c 

I 

1. 

q        p        s 
r        s       p 

2. 

1      y      I 

1           Z           C 

• 

p   -q        r 

0  -q   -r 

3. 

q      P  -  s 
-r        s       p 

4. 

q        0   -  s 
r        s       C 

Find  the  values  of  the  following  determinants. 

5. 

1  2        3 
3        1        2 

2  3        1 

6. 

1-3       4 

2        0-5 
3-1        7 

7. 

8       9 
2       3 
1        1 
4       3 

0 
0 
6 
5 

tti  a2  fls 

«! 

bi  Cl 

bi  ai  Cl 

h  h  63 

= 

ffl2    b'Z    Cl 

=  - 

62  a2  C2 

Cl     C2     C3 

as  63  C3 

bs  as  C3 

«!   32   as 
61   62   bs 

Cl     Co     C3 

= 

fll 

&2    63 
C2    C3 

-as 

bi  63 

Cl    C3 

Prove  the  following  relations  by  expanding  the  determinants. 


\bi  62 1 


10.  In  the  expansion  of  the  determinant  |  ai  62  C3  ^4 1  collect  all  the 
terms  which  involve  as  factors  (1)  Csdi,  (2)  01^4,  (3)  a263di,  (4)  ai,  (5)  C3. 

11.  Find  the  signs  of  the  following  terms  in  the  expansion  of  the 
determinant  \ai  62  cs  di  e^]. 

a2biC3die5.  a^boCidnes.  a^biCsd^ei. 

01^2036465.  Cibzesaids.  dsOzeibiC^. 


498 


A   COLLEGE   ALGEBRA 


899 


900 


901 


902 


PROPERTIES    OF   DETERMINANTS 

Theorem  1.  The  value  of  a  determinant  is  not  chayiged  if  its 
rows  are  made  columns,  and  -its  columns  rows,  without  changing 
their  relative  order. 


Thus, 


«!      flo      f'o 

a  I  l>i  Ci 

h,  h,  h. 

Go     bn     Co 

^1      Co     Cs 

(1) 

^3   h   Cz 

(2) 


For  every  term,  as  aJb^Ci,  in  the  expansion  of  the  deter- 
minant marked  (1)  contains  one  element  and  but  one  from 
each  row  and  column  of  (1).  Hence  it  contains  one  element 
and  but  one  from  each  column  and  row  of  (2)  and  is  therefore, 
except  perhaps  for  sign,  a  term  of  (2).  Moreover  the  sign  of 
the  term  in  (1)  is  the  same  as  its  sign  in  (2).  For  if  the 
factors  of  the  term  be  arranged  in  the  order  of  the  letters 
a,  b,  c,  the  inversions  of  the  subscripts  will  determine  this  sign 
in  both  (1)  and  (2),  in  (1)  by  the  rule  of  §  890,  in  (2)  by  the 
rule  of  §  891. 

Conversely,  every  term  in  the  expansion  of  (2)  is  a  term 
in  the  expansion  of  (1). 

Hence  for  every  theorem  respecting  the  rows  of  a  determi- 
nant there  is  a  corresponding  theorem  respecting  its  columns. 

Theorem  2.  If  nil  the  elements  of  a.  row  or  column  of  a  deter- 
minant are  0,  the  value  of  the  determinant  is  0. 

For  every  term  of  the  determinant  Avill  contain  a  factor 
from  the  row  or  column  in  question,  §  890,  and  will  therefore 
vanish. 

Theorem  3.  If  two  rows  (or  columns)  of  a  determinant  be 
interchanged,  the  determinant  merely  changes  its  sign. 


Thus, 


"l     Oo     «3 

^1      ^2      ^3 

bi    h  h 

=  — 

b,     b,     b. 

Ci     C2     Cg 

(1) 

rtj  a^  «3 

(2) 


DETERMINANTS   AND   ELIMINATION 


499 


Yov  any  term  of  (1)  with  factors  arranged  in  the  order  of 
the  rows  of  (1)  may  be  transformed  into  a  term  of  (2)  with 
factors  arranged  in  the  order  of  the  rows  of  (2)  by  inter- 
changing its  first  and  last  factors ;  and  conversely.  But  this 
interchange  will  increase  or  diminish  the  inversions  of  the 
subscripts  in  the  term  by  an  odd  number,  §  887,  and  therefore, 
since  the  normal  order  of  the  subscripts  is  123  in  both  (1)  and 
(2),  it  will  change  the  sign  of  the  term. 

Thus,  a2b3Ci  is  a  term  of  (1)  and  —  cibsU'z  is  the  corresponding  term 
of  (2).  For  in  ao&sCi  the  subscripts  present  two  inversions,  while  in 
Cibsa-z  they  present  one  inversion. 

Example.  Verify  the  preceding  theorem  by  expanding  each  of  the 
determinants  (1)  and  (2). 

Corollary.     If  two  of  the  roivs  (or  columns)  of  a  determinant     903 
are  identical,  the  deterniinant  vanishes. 

For  let  D  denote  the  value  of  the  determinant.  An  inter- 
change of  the  identical  rows  must  leave  D  unchanged ;  but,  by 
§  902,  it  will  change  D  into  —  D. 

Therefore  D^—  D,  that  is,  2  D  =  0,  ot  D  =  0. 

a  a  d 
Thus,        b  b  e    =  abf  +  aec  +  dbc  —  aec  —  abf  —  dbc  =  0. 
c  c  f 

Theorem  4.     If  all  the  elements  of  a  row  (or  column)  are     904 
multiplied  by  the  same  number,  as  k,  the  determinant  is  multi' 
plied  by  k. 

For  of  the  elements  thus  multiplied  by  k  one  and  but  one 
occurs  as  a  factor  in  each  term  of  the  determinant,  §  890. 

The  evaluation  of  a  determinant  may  often  be  simplified  by 
applying  this  theorem. 
Thus, 


=  480. 


8 

2 

-3 

4 

1 

-1 

1 

1 

>0 

5 

=  2-5- 

3 

-4 

1 

=  2.5.3.4- 

1  - 

-1 

1 

4 

-1 

3 

4 

-1 

1 

1 

-1 

500 


A    COLLEGE    ALGEBRA 


905 


906 


907 


Corollary.     If  the  corresponding  elements  of  two  columns  (or 
roivs)  are  proportional,  the  determinant  vanishes. 


Thus, 


ra  a  d 

a  a  d 

rb   b   e 

=  r 

b   b   e 

re    c  f 

c   e  f 

?•  •  0  =  0. 


§§  903,  904 


Theorem  5.  If  one  of  the  columns  (or  roivs)  has  binomial 
elements,  the  determinant  may  be  expressed  as  a  sum  of  two 
determinants  tw  the  manner  ilhistrated  below. 


«i  +  a' 
b,  +b' 
Ci  +  c' 


«2    «s 
b,   b, 

Co     Co 


«1     «2     «3 

a'  02  as 

= 

bi   b^   bs 

+ 

b'   b,    b. 

0) 

Cl      ^2      ^3 

(2) 

C'    Co     ^3 

(3) 


For  each  term  of  (1)  is  the  sum  of  the  corresponding  terms 
of  (2)  and  (3).      Thus,  (aj  +  a')boCz  =  aib^Cs  +  a'bnC^. 
Observe  that  any  of  the  numbers  a',  b',  c'  may  be  0. 

By  repeated  applications  of  this  theorem  any  determinant 
with  polynomial  elements  may  be  resolved  into  a  sum  of 
determinants  with  simple  elements. 

ai  +  a'l      02  +  a' 2      03  +  a'z 

Example.     Express    the    determinant     hy  +  h\      60  +  h'n      63  +  V^ 
as  a  sum  of  eight  determinants.  Ci  +  c\      Cn  +  c'2      C3  +  c'3 

Theorem  6.  The  vahie  of  a  determinayit  tvill  not  be  changed 
if  to  the  elements  of  any  column  (or  row)  there  be  added  the 
corresjjondi^ig  elements  of  any  other  column  (or  row)  all  multi- 
pflied  by  the  same  number,  as  k. 

Thus,  by  the  theorems  of  §§  905,  906,  we  have 

ai  +  ka^  a. 

bi  +  kb^  b^ 

Cl  +  kcz   Co 
And  similarly  in  any  other  case. 

It  follows  from  this  theorem  that  a  determinant  vanishes  if 
one  of  its  rows  may  be  obtained  by  adding  multiples  of  any 
of  its  remaining  rows. 


flz 

'h   cio   f/3 

h 

= 

h,    b,    b. 

+ 

C3 

<-l     (-2     C, 

kaa  a^  a^ 

a-i  a<2,  dz 

kbg    ^2    ba 

= 

Z»i    b^   bs 

kc^    C2    C3 

Cl    C2    Cs 

DETERMINANTS   AND    ELIMINATION 


501 


Thus, 


4  7  7 
5-4  2 
2        5       1 


0  since  4,  7,  7  =  2(5,  -  4,  2)  +  3(-  2,  5,  1). 


Theorem  7.  //"a  determinant  whose  elements  are  rational 
integral  functions  of  some  variable,  as  x,  vanishes  when  x  =  a, 
the  determinant  is  divisible  by  x  —  a,. 

For  the  determinant  when  expanded  may  be  reduced  to  the 
form  of  a  polynomial  in  x.  And  since  this  polynomial  vanishes 
when  ic  =  a  it  is  divisible  by  a;  —  a,  §  415. 

The  factors  of  a  determinant  may  often  be  found  by  aid  of 
this  theorem. 


Example.     Show  that 


1 

1 

1 

a 

b 

c 

a- 

62 

C2 

:{a  —  b){b  —  c)  (c  —  a). 


By  §  903,  this  determinant  will  vanish  if  a  =  6,  if  6  =  c,  or  if  c  =  a. 
Hence  it  is  divisible  by  a  —  6,  b  ~  c,  and  c  —  a,  ^  416,  and  therefore  by 
the  product  (a  —  b){b  —  c)  (c  —  a).  But  this  product  and  the  determinant 
itself  are  of  the  same  degree,  three,  with  respect  to  a,  b,  c.  Hence  the 
two  will  differ  by  a  numerical  factor  at  most. 

One  term  of  the  determinant  is  ftc^  and  the  corresponding  term  of 
{a  —  b){b  —  c)  (c  —  a)  is  bc"^.  Hence  tlie  numerical  factor  in  question  is  1, 
and  the  determinant  is  equal  to  (a  —  b)  {b  —  c)  (c  —  a). 


908 


EXERCISE   LXXXII 
1.    Evaluate  the  following  determinants. 


(1) 


6 

42 

27 

10 

8 

-28 

36 

•         (2) 

15 

20 

35 

135 

20 

2.    Prove  that 


ffli  +  fcaz  + 


Cl  +  kC2  +  Ics 


8       2 

-ab 

ac 

ae 

12       3 

(3) 

bd  -cd 

de 

32      12 

bf       cf 

-ef 

a.2  +  mQLs     as 

ai  ao  as 

&2  +  mbs     ba 

6i  62  63 

Cl  +  mcs 

C3 

c 

1   C2  C3 

3.    By  aid  of  the  theorem  of  §  907  prove  that  the  value  of  each  of  the 
foUowins;  determinants  is  0. 


(1) 


a  d 

c  d 

c  b 

a  b 


(2) 


1 

p 

5 

1 

Q 

r 

1 

r 

s 

1 

s 

P 

r  +  s 
p  +  s 

p  +  q 

q  +  r 


(3) 


2  1  4 
3-1  2 
1        2       3 


-  1 

-  1 
-2 
-2 


50i 


A    COLLEGE   ALGEBRA 


4.    Prove  that 


5.    Prove  that 


1  p  p3 
1  q  q^ 
1       r       r3 


iP  -q){q-r){r-p){p  +  q  +  r). 


{b  +  c)2  a6  oc 

ab  (c  +  a)2  6c 

ac  be  {a  +  6)2 


2abc{a  +  b  +  c)3. 


909 


MINORS.     MULTIPLICATION   OF   DETERMINANTS 

Minors.  In  any  determinant  A  suppress  the  row  and  column 
in  which  some  particular  element  e  lies  and  then  form  the 
determinant  of  the  remaining  elements  without  disturbing 
their  relative  positions.  This  new  determinant  is  called  the 
complementary  minor  of  the  element  e  in  question  and  may 
be  denoted  by  A^. 


Thus,  in  A 


cti 

02 

aa 

bx 

62 

&3 

Cl 

C2 

C3 

the  minor  of  Cj  is  Ac 


fl2    <^3 

62     63 


910  Theorem.     In  the  expansion  of  any  determinant  A  the  stun  of 

all  the  terms  ivhich  involve  the  leading  element  ai  is  aiA  . 


Thus,  in 


«! 

a^ 

a  3 

«4 

h 

h 

^•3 

^4 

Cl 

C-l 

Cz 

Ci 

di 

ch 

dz 

d. 

this  sum  is  Oi 


(1) 


h   h   h 

C„        Cg       C4 

d.  dz  di 


(2) 


For,  apart  from  sign,  each  term  of  A  which  involves  a^  is 
formed  by  multiplying  a^  by  elements  chosen  from  the  remain- 
ing rows  and  columns  of  A,  one  from  each,  in  other  words,  by 
multiplying  a^  by  a  term  of  A„j.  Moreover  the  sign  of  the 
A  term  is  that  of  this  A,,^  term,  since  writing  rtj  before  the  latter 
term  will  not  affect  the  inversions  of  the  subscripts.  Thus, 
the  term  —  aib^r^d^  of  (1)  may  be  formed  by  multiplying  «! 
by  the  term  —  biC^d^  of  (2).  Conversely,  the  product  of  ai  by 
each  term  of  A„  is  a  term  of  A. 


DETERMIXANTS   AXD   ELIMINATIOX 


503 


Corollary.     If  e  denotes  the  element  in  the  ith  roiv  and  Vth     911 
column  of  A,  the  sum  of  all  the  terms  of  A  which  involve  e  is 
(-ly  +  '^eA,. 

For  we  can  bring  e  to  the  position  of  leading  element  without 
disturbing  the  relative  positions  of  the  elements  which  lie 
outside  of  the  row  and  column  in  which  e  stands,  namely,  by 
first  interchanging  the  row  in  which  e  stands  with  each  pre- 
ceding row  in  turn  and  then  interchanging  the  column  in 
which  e  stands  with  each  preceding  column.  In  carrying  out 
these  successive  interchanges  of  rows  and  columns  we  merely 
change  the  sign  of  the  determinant  (i  —  V)-\-(k  —  l)ovi-\-k  —  2 
times,  §  902.  Hence,  if  A'  denote  the  determinant  in  its  final 
form,  we  shall  have 


A'  =  (-l)'  +  ^ 


(-  ly-^A. 


By  §  910,  the  sum  of  all  the  terms  in  A'  which  involve  e  is 
eA'^.  Hence  in  A  this  sum  is  (—  l)'"'"*-eA^.  For  the  minor  of 
e  in  A  is  the  same  as  its  minor  in  A'. 

Thus,  in  the  case  of  A  =  |  ai  62  C3  ^4 1  the  element  ds,  for  which  i  =  4, 
fc  =  3,  may  be  brought  to  the  leading  position  as  follows  : 


ax   02  as  ai 

h   62  63  64 

Cl  C.2     C3  C4 

di  di  dz  di 

(1) 

di   ^2  ds  di 

«i  «2  «3  ai 

61  62  63  &4 

Cl  C2  Cs  Ci 

(2) 

da  di 

do  d4 

as  ai 

02  ai 

h  h 

62  64 

C3     Cl 

C2  C4 

(3) 

By  interchanging  the  fourth  row  of  (1)  with  the  third,  second,  and 
first  in  turn,  we  obtain  (2)  before  which  we  place  the  minus  sign  because 
of  the  three,  that  is,  i  —  1,  interchanges  of  rows. 

Then  by  interchanging  the  third  column  of  (2)  with  the  second  and 
first  in  turn,  we  obtain  (3)  before  which  we  place  the  same  sign  as  that 
before  (2)  because  of  the  two,  that  is,  A;  —  1,  interchanges  of  columns. 

The  minor  of  ds  in  (1)  is  the  same  as  its  minor  in  (3).  Hence  the  sum 
of  all  the  terms  of  (1)  which  involve  da  is  —  ds  •  |  Oi  bo  C4 1. 

Theorei6.    A  determinant  may  he  expressed  as  the  sum  of  the     913 
products  of  the  elements  of  07ie  of  its  7'ows  or  columns  by  their 


504  A    COLLEGE   ALGEBRA 

comijlementary  minors,   ivlth  signs  which  are  alternately  plus 
and  minus,  or  minus  and  plus. 

Thus,  in  the  case  of  a  determinant  of  the  fourth  order 

A  =  I «!  ^2  Cg  di  I  we  have 

A  =  aiA„^  -  «2A„^  +  a3A„3  -  ai\^. 

For  each  term  in  the  expansion  of  A  contains  one  and  but 
one  of  the  elements  a,,  a^,  a^,  04.  And,  by  §§  910,  911,  the 
sum  of  all  the  terms  which  involve  a^  is  a^A^ ,  the  sum  of  all 
which  involve  a„  is  —  a^^^,^,  and  so  on. 

In  like  manner, 

A  =  -  b,\,  +  h\,  -  h\,  +  h\ 

=  aiA„j  —  JjAjj  +  CiA^j  —  diAj^,  and  so  on. 

913         Cofactors.     It   is   sometimes   more  convenient  to  write  the 
preceding  expressions  for  A  in  the  form 

A  =  a^Ai   +  ^2.12  +   "3-I3  +   «4^4 

=  b^B,  +  boB.  +  bsBs  +  b,Bi, 

and  so  on,  where  Ai  =  A„^,  A2=—  A„^,  and  so  on.     We  then 
call  A  I,  A2,  ■  ■  ■  the  cofactors  of  ai,  a^,  ■  •  ■. 
Thus,  in 


ai 

a2 

as 

61 

h 

h 

Cl 

cz 

C3 

the  cofactors  of  ai,  On,  as  are 


1^2  C3 1'        |ci  Car    |ci  Co 


914  -An//  sum  like  biAi  +  b2A2  +  bjAg  +  b4A4,  obtained  by  adding 

the  products  of  the  elements  of  one  row  by  the  cofactors  of  the 
corresponding  elements  of  another  roiv,  is  zero. 

For  b^Ai  +  />2.l2  +  ^3-43  +  ^i^i  denotes  a  determinant  whose 
last  three  rows  are  the  same  as  those  of  A  =|ai  b^,  c^  di\  but 
whose  first  row  is  b^,  b^,  b^,  b^.  And  since  both  the  first  and 
second  rows  of  this  determinant  are  bi,  b^,  b^,  b^,  it  vanishes, 
§  903.     And  so  in  general. 


DETERMINANTS    AND   ELIMINATION 


605 


Bordering  a  determinant.     Any  determinant  may  be  expressed     915 
as  a  determinant  of  a  higher  order.     For,  by  §  912,  we  have 


«!  aa  ^3 

W   h   ^3 

= 

^1      ''2      CS 

10  0 

0  fti  a^ 

0  b,  b, 

0    Ci  C2 


0 


,  and  so  on. 


Evaluation  of  a  determinant.  The  value  of  a  numerical  deter- 
minant of  any  order  may  be  obtained  by  aid  of  the  theorem  of 
§  912  and  the  rule  of  §  898 


916 


Thus, 


2 

3 

1 

-1 

2       0       0       3 

2 

0 

0 

3 

2        3        1-1 

4 

1 

0 

1 

•~ 

4        10       1 

1 

2 

—  2 

1 

-1        2-2        1 

3        1-1 

2 

3 

1! 

=  —  2 

1  0       1 

2  -2       1 

+  3 

4 
-1 

1 
2 

^1 

-21 

=  — 

le 

+  87  =  G9. 

But  in  most  cases  the  value  of  a  numerical  determinant  may 
be  found  with  less  reckoning  by  the  following  method. 
It.  follows  from  §§  904,  907,  910  that 


«1     «2    tts     •  •  • 

(ii  GiCt^  (iiao  •  •• 

bi     b.^     Z>3     ••• 

Ci    c-a    Cs    ••• 

(1) 

1 

<'i    «i<'2    «i^3    ••• 

(2) 

1 

a^bo  —  a«bi  a^b^  —  ajby  •  •  ■ 

^^ 

a^c^  —  aa^i   a^Cs  —  a^Ci   •  ■  • 

1 

(3) 

Here  (1)  denotes  a  determinant  of  the  7ztli  order.  By  multi- 
plying each  column  of  (1)  after  the  first  by  «!  we  obtain  (2). 
From  the  second  column  of  (2)  we  subtract  the  first  multiplied 
by  02 ;  from  the  third  column  we  subtract  the  first  multiplied 
by  ftg ;  and  so  on.     The  first  row  of  the  resulting  determinant 


917 


506 


A    COLLEGE   ALGEBRA 


is  «!,  0,  0,  •  •  •.  Hence  this  determinant  is  equal  to  a^  times  its 
minor,  which  is  the  determinant  of  the  (n  —  l)th  order  (3). 

Observe  that  each  element  of  (3)  is  obtained  from  the  cor- 
responding element  of  the  minor  of  a^  in  (1)  by  multiplying 
that  element  by  «i  and  from  the  result  subtracting  the  product 
of  the  corresponding  elements  in  the  first  row  and  the  first 
column  of  (1). 

Thus,  by  two  reductions  of  the  kind  just  described,  we  have 


2       2-1        3 
2        13-2 
2-121 
3-2-2        1 

_  1 

~22 

6 
-    6 
-10 

4       2 

6   -4 

-1    -7 

= 

3       2       1 

3-3       2 
10       1        7 

1     -15 
~3     -17 

.?=--. 

2-  1  -  2(-  2)  = 

-6,  2 

3-( 

-l)(-2 

= 

4,  and  so  on. 

When  the  leading  element  is  0  ^Ye  begin  by  bringing  another  element 
into  the  leading  position  by  the  method  explained  in  §  911. 

Example.     Evaluate  each  of  the  following  determinants  by  both  of  the 
methods  just  described. 

1-2       3       4 


1  3 

0  -  1 

2  3 

1  -3 


918  Multiplication  of  determinants.  The  product  of  two  deter- 
minants of  the  same  order,  A  and  A',  may  be  expressed  in  the 
form  of  a  third  determinant  A"  obtained  as  follows  : 

Multiply  the  elements  of  the  \th  row  of  A  hy  the  corresponding 
elements  of  the  \.th  row  of  A'.  The  sum  of  the  products  thus 
obtained  is  the  element  in  the  \th  row  and  \th  column  of  A". 


Thus, 


«1     ^2 


\<lx   lA 


For,  by  §  906,  the  third  determinant  is  the  sum  of 

klPl     ai-Zll  W\Pl     «272l  WlPl    «l?l|  |«2/'2     «22'2l 

\b\P\    ^1^1 1(1)       V>xP\  *2?3r(2)       \biPi   %ir(3)        \\p%   %2|(4) 


DETERMINANTS   AND    ELIMINATION 


507 


But  (1)  and  (4)  vanish,  since  their  columns  are  proportional, 
§  905.  And  simplifying  (2)  and  (3)  by  aid  of  §§  902,  904,  and 
adding  them,  we  have 


i'i?2 


h   h 


^Pi<l\ 


{vx<ii  -  P2qi) 


\Pi  P^ 

\l\     ^2  1 


bi   h 


«1     «2 

k  ^2 


«1     «2     «3 

Pi  Pi  Pz 

bi     ^2     ^3 

q\  ?2  qz 

^1      ^2      ^3 

ri   r^   n 

Again, 


(f'lPx  +  «2j32  +  a^Pz         ai?!  +  «2?2  +  azqz         «'in  +  «2^2  +  ciz>'z 
hPi.   +  hPi   +  ^ZPZ  \q\    +  *2?2   +  *3?3  ^>1^1   +  ^2^2  +  ^3^3 


as  may  be  shown  by  resolving  the  third  determinant  into  a 
sum  of  determinants  with  simple  columns  in  the  manner  just 
illustrated.  There  will  be  twenty-seven  such  determinants, 
but  twenty -one  of  them  have  two  or  three  proportional  columns, 
and  therefore  vanish.  Each  of  the  remaining  six  is  equal  to 
the  determinant  |ai  h.^,  Cz\  multiplied  by  a  term  of  |pi  q^,  rgj, 
so  that  their  sum  is  \a^  b^  Cz\-  \pi  q^  r^\. 
This  proof  may  easily  be  generalized. 

The  rule  above  given  is  readily  extended  to  determinants 
of  different  orders.  We  have  only  to  begin  by  making  their 
orders  the  same  by  bordering  the  one  of  lower  order,  §  915. 


EXERCISE   LXXXIII 


Evaluate  the  following  determinants. 


/ 


2  -1 

7  5 

-3  2 

4  7 


/ 


1 

2 

3 

-  1 

-2 

2 

1 

3 

-2 

1 

0 

1 

2 

_  2 

1 

0 

-1 

-1 

2 

1 

0 

2 

3 

1 

-1 

508 


A   COLLEGE   ALGEBRA 


12 


6 

-    4 

10 

26 

18 

6 

-30 

21 

12 

24 

40 

28 

9 

-    2 

20 

14 

Express  each  of  the  following  products  as  a  determinant. 


a 

b 

c 

b 

c 

a 

c 

a 

b 

a  —  a  a 

b  b  b 

c  c  —  c 

d  d  d 


a 

b 

a  b 

I    m  n 

'    c  d  ' 

8. 

m  n   I 

c 

d 

n    I     ni 

«1    «2     03 

bi  b-2   bs 

Ci    C.2     Cs 

Ax  A2  A3 

ai  fflj  «3 

B,  B,   B3 

— 

61  b,  63 

Ci   C,    C3 

Cl     C2    C3 

p     0     r 

a    0    c 

6. 

p     q     0 

a    b    0 

0     q     r 

0    b    c 

9.  Prove  that 


10.    Prove  that  a  determinant  reduces  to  its  leading  term  when  all  of 
the  elements  at  either  side  of  the  leading  diagonal  are  zero. 


919 


ELIMINATION.     LINEAR   EQUATIONS 

Solution  of  a  system  of  linear  equations.     We  are  to  solve  the 
following  system  of  equations  for  x^,  x^,  X3 : 
a^Xi  +  a^x^  +  CT3X3  =  k  1 

bix^  +  b^x^  +  63X3  =  I    I.  (1) 

c^xi  +  C2X2  +  C3X3  =  mj 

Let  A  denote  |ffi  b^  03],  the  determinant  of  the  coefficients 
of  Xi,  x^,  X3  arranged  as  in  (1),  and,  as  in  §  913,  let  A^,  A^,  ■  ■  • 
denote  the  cofactors  of  a^,  a^,  •  •  •  in  A. 

To  eliminate  x^  and  Xg,  we  multiply  the  first  equation  by  Ai, 
the  second  by  B^,  the  third  by  Cj,  and  add.     We  obtain 
aiAi  Xi  +  a.^Ai  .r,  +  a^Ai  X3  =  kA^ 
h,lh        +h,B,        +h3B,        +//>', 
^i^'i         +C2<\         +<'3<\        +m(\ 
But  in  this   equation  the  coefficients  of  x.,  and  X3  are  0, 
§  914  ;  the  coefficient  of  x^  is  A,  §  913  j  and  the  secord  member 


DETERMINANTS   AND   ELIMINATION 


509 


denotes  a  determinant  found  by  replacing  the  first  column  of 
A  by  k,  I,  m.     Hence  the  equation  may  be  written 


(2) 


We  may  in  like  manner  derive  an  equation  which  involves 
a-2  alone  by  multiplying  the  given  equations  by  ylj,  B^,  C\ 
respectively  and  then  adding  them,  and  an  equation  which 
involves  x^  alone  by  multiplying  by  .43,  B^,  C\  and  adding. 
These  equations  are 


«1     0^2     «3 

k    a^  as 

b,     b,     b. 

J'l  = 

I     b,    b. 

Ci    c^    c. 

m  ("2    fg 

«!     CU     (Iz 

«1 

k    as 

b,  b,  b. 

^1 

I        bs 

Ci     Co     ('3 

<^l 

m  Cs 

(3), 


Hence,  if  A  ^  0,  the  required  solution  is 


»l 

«2     «3 

ai  a.j.  k 

1>1 

b,     bs 

Xs  = 

b^    b^    I 

<h 

Co     Cs 

Ci    C2    m, 

(4). 


a-i 


\k    b,  c,\ 
Irt,   bo  Co\ 


\<h  I     ^3 1 


l^i  bo,  m\ 


l«i  ^2  ^3!  Vh  ^2  Cs\ 

that  is,  the  value  of  each  unknown  letter  x^,  x^,  x^  may  be 
expressed  as  a  fraction  whose  denominator  is  A  and  whose 
numerator  is  a  determinant  differing  from  A  only  in  this,  that 
the  coefficients  of  the  unknown  letter  in  question  are  replaced 
by  the  known  terms. 

It  may  be  proved  in  the  same  way  that  the  like  is  true  of  a 
system  of  n  linear  equations  in  n  unknown  letters. 

Example.     Solve  2x  —  3y  +  2  =  4, 

X  +  2/  —  z  =  2, 

i.x  —  y  +  Zz  =  \. 


"We  have 


In  like  manner,  we  find  y 


2  -S 
1  1 
4   -  1 

21/20, 


"i_26 


-35/20  =  -7/4. 


If  A  is  0  and  any  one  of  the  determinants  |  /^  ^2  ^sl?  |  «i  I  Cs\, 
I «!  />2  m  I  is  not  0,  it  follows  from  (2),  (3),  (4)  that  the  given 
equations  (1)  have  no  finite  solution  (compare  §  394). 


920 


510  A   COLLEGE   ALGEBRA 

If  A  and  all  the  determinants  \k  b^  ^3],  |ai  I  Cg],  |ai  Jg  ^1 
vanish,  the  equations  (2),  (3),  (4)  impose  no  restriction  on  the 
values  of  x-^,  x^,  x^.  In  this  case  the  given  equations  (1)  are  not 
independent.  This  follows,  by  §  394,  from  the  manner  in 
which  (2),  (3),  (4)  were  derived  from  (1),  unless  all  the 
minors  A^,  An,  •••  vanish.  And  if  all  the  minors  vanish,  it 
may  readily  be  shown  that  the  three  equations  (1)  differ  only 
by  constant  factors,  so  that  every  solution  of  one  of  them  is  a 
solution  of  the  other  two. 

These  results  are  readily  generalized  for  a  system  of  n  equa- 
tions in  71  unknown  letters. 
921         Homogeneous   linear    equations.     When    k  =  I  =  rn  =  0,    the 
equations  (1)  of  §  919  reduce  to  a  system  of  homogeneous 
equations  in  x^,  x^,  Xs,  namely, 

ajXi  +  a^x^  +  «33:"3  =  Q 1 

b,})pi  +  hx,  +  hx,=il,  (1) 

CiXi  +  C2X2  +  CsXs  =  e)  J  . 

and  the  equations  (2),  (3),  (4)  of  §  919  become 

Aa-i  =  0,  Ax^  =  0,  Axs  =  0.  (2) 

Evidently  the  equations  (1)  have  the  solution  Xj  =  a'g  =  Xg  =  0, 
and  it  follows  from  (2)  that  this  is  the  only  solution  unless 
A  =  0. 

But  if  A  =  0,  the  equations  (1)  are  satisfied  by 

Xi  =  ?vli,  Xo  =  rAo,  X3  =  rAz,  (3) 

where  r  may  denote  any  constant  whatsoever. 

For,  substituting  these  values  in  (1)  and  simplifying,  we  have 

a^Ai  +  a^A^  +  «3'43  =  0,  b^A^  +  /^2.42  +  ^3^4 3  =  0> 
CiAi  +  C2A2  +  ^3^43  =  0, 

and  these  are  true  identities,  the  first  one  because  A  =  0,  the 
other  two  by  §  914.  The  same  thing  may  be  proved  as  follows  : 
If  we  solve  the  second  and  third  of  the  equations  (1)  for  Xi 
and  X.2  in  terms  of  x^,  we  obtain  x^/  A^  =  x^/Ao  =  x^/ A^,  or,  if 


DETERMINANTS    AND    ELBIINATION  511 

r  denote  the  value  of  these  equal  ratios,  x^  =  rA^,  x^  =  rA^, 
Xs  =  rA-i.  And  as  just  shown,  if  A  =  0,  these  values  will  also 
satisfy  the  first  of  the  equations  (1). 

From  this  second  proof  it  follows  that  when  A  =  0 
Xi :  Xz'.  Xs  =  Ai_ :  A^:  A  3  =  Bi :  B2  :  Bs  =  Ci :  C\  :  C\, 

that  is,  the  minors  of  corresponding  elements  in  the  rows  of 
A  are  proportional.  It  is  assumed  that  these  minors  are  not  0. 
From  the  system  of  three  non-homogeneous  equations  in  x,  y     922 

a^x  +  a.^ij  +  ag  =  0 1 

M  +  %  +  ^>3  =  0l,  (1') 

CiX   +  C.JJ   +  (-3   =  0  J 

we  may  derive  the  homogeneous  system  (1)  of  §  921  by  sub- 
stituting X  =  Xi/xg,  y  =  x^/x^  and  clearing  of  fractions. 

Hence  A  =  0  is  the  condition  that  the  equations  (1')  have  a 
common  solution. 

EXERCISE   LXXXIV 

Solve  the  following  systems  of  equations  by  determinants. 

r2x  +  3?/-5z  =  3,  r2x-H4?/-32  =  3,  '' 

1.    -!  X  -  2?/ +  z  =  0,  ^  2.    <:  3x- 8y +  6z  =  1, 

l3x  +  7/  +  3z  =  7.  ^  l8x-2?/-0z  =  4. 

r2x-4y  +  3z  +  4«  =  -3, 
(  az  +  by  +  cz  =  a, 
I  '  3x-2y-h6z  +  5«  =  -l, 

I  I  5x  +  8?/  +  f)z  +  3«  =  9, 

[a^x  +  ¥y  +  c^z  =  d^  nn         <,         -.0 

Lx  -  lOy  -  3z  -  t  t  =  2. 

Show  that  the  following  systems  of  equations  are  consistent,  and  solve 
them  for  the  ratios  x-.y-.z. 

c  X  +  2  y  -  z  =  0,  fciix  +  biy  +  (kcii  +  Ibi)  z  =.  0, 

5.    \sx-y  +  Az  =  0,  6.    J  aox  +  h.y  +  (ka2  +  lb..)  z  =  0, 

L 4 X  +  2/  +  3  z  =  0.  [asx  +  bsy  +  {has  +  ^^3)  z  =  0. 

7 .    For  what  values  of  X  are  the  following  equations  consistent  ? 
4  X  +  3  y  +  z  =  Xx, 
3x  —  iy  +  1  z  =  \y, 
x  +1  y  -6z  =  \z. 


512 


A    COLLEGE   ALGEBRA 


RESULTANTS 


923  Resultants.  By  the  resultant  of  two  equations  f{x)  =  0  and 
(^  (x)  =  0  is  meant  that  integral  function  of  the  coefficients  of 
f(x)  and  (f)  (x)  whose  vanishing  is  the  necessary  and  sufficient 
condition  that/(x)  =  0  and  (f)(x)=  0  have  a  common  root. 

Thus,  the  resultant  of  a^x,^  +  aiX.  +  a^  =  0  (1)  and  x  —  b  =  0  (2)  is 
aob"^  +  aib  +  a^ ;  for  when  aolfi  -f  a^b  +  ao  =  0,  the  equations  (1)  and  (2) 
have  the  common  root  b. 

924  The  resultant  of  any  two  equations  /(a-)  =  0,  <^  (x)  =  0  may 
be  obtained  by  eliminating  x  by  the  following  method  due  to 
Sylvester. 

To  fix  the  ideas,  let 

f{x)  =  Qox'^  -f  a^x"^  +  a^x  +  a 3  =  0,  (1) 

<!>  (x)  =  b,p:'  +  hx  +  b,  =  0.  (2) 

Multiply  (1)  by  x  and  1,  and  (2)  by  x^,  x,  and  1  successively. 

We  obtain  4,3,2,  n 

a^jx*  -\-  a^x^^  -\-  a<pu^  +  a^x  =  0, 

«(iX^  +  O'i^'  +  a^x  +  ag  =  0, 

h^x"  +  hix^  +  h.^""  =  0, 

bfpc^  +  b^x^  +  b^x  =  0, 

b.x''  +  &1.X  +  b.  =  0. 

These  may  be  regarded  as  a  system  of  five  homogeneous 
linear  equations  in  the  five  quantities  x*,  x^,  x"^,  x,  1.  Hence, 
§  921,  they  cannot  have  a  common  solution  unless 


0    a, 

0  b, 
0  0   b 


»2 

«.3 

0 

a  I 

O2 

«3 

h 

0 

0 

f'l 

b. 

0 

h) 

h 

^'2 

(3) 


Hence  (3)  is  the  necessanj  condition  that  (1)  and  (2)  have  a 
common  root.  It  is  also  the  suffirlent  condition.  For  to  the 
fifth  column  of  D  add  the  first  four  columns  multiplied  by  a;*, 


DETERMINANTS    AND    ELIMINATION  513 

,r',  x^,  X  respectively.  We  thus  transform  D  into  an  equivalent 
determinant,  §  907,  whose  fifth  column  has  the  elements  ^f(x), 
f(x),  x^<f>  (x),  .T<^  (jc),  cf)  (x).  Hence,  if  fx-i,  1x2,  h-s,  H-i^  H-b  denote  the 
cofactors  of  the  elements  of  the  fifth  column  of  D,  we  have, 

§  ^^^'    D  =  (fJi.X  +  fJi.)f(x)  +  (/X3.f'^  +  f,,X  +  ix,)  </>  (x)  =  0. 

It  follows  from  this  identity  that  each  factor  a-  —  /8  of  f(x) 
must  be  a  factor  of  (fx^x^  +  fiiX  +  /xg)  ^(a:*),  and  therefore,  since 
f(x)  is  of  the  third  degree  and  /xsX^  +  fXiX  +  /x^  is  of  only  the 
second  degree,  that  at  least  one  factor  x  —  ^  oif(x)  must  be 
a  factor  of  f^{x),  in  other  words,  that  one  of  the  roots  of 
f(x)  =  0  must  be  a  root  of  <^(x)=  0,  §  795. 

It  is  here  assumed  that  the  minors  fxi,  fx^,  •■-,  fx^  are  not  all  zero. 
If  the  minors  of  all  the  elements  of  D  are  0,  it  can  be  proved 
that/'(.r)  =  0  and  <^(a')  =  0  have  more  than  one  common  root. 

If  Xi,  A2,  •  •  •,  A5  denote  the  cofactors  of  the  elements  of  any     925 
row  of  D,  it  follows  from  §  921  that  when  D  =  0 

a-*  :  x^  :  ic^ :  a- :  1  =  Xi :  X2 :  A3 :  A4 :  A5, 

whence  x  =  Ai/Ao  =  A2/A3  =  •  •  •  =  A4/A5.  Therefore  when 
f(x)  =  0  and  (f>  (a-)  =  0  have  a  common  root,  the  value  of  this 
root  is  Ai/Ao. 

In  the  general  case  when  the  degrees  of  f(x)  =  0  and  926 
(f)  (x)  =  0  are  m  and  u  respectively,  the  resultant  D  will  be  a 
determinant  of  the  (w  +  ^>)th  order  whose  first  n  rows  consist 
of  the  coefficients  of  f(x)  and  zeros  and  whose  remaining  m 
rows  consist  of  the  coefficients  of  ^(x)  and  zeros,  arranged 
as  in  §  924,  (3).  Hence  in  the  terms  of  D  the  coefficients  of 
f(x)  enter  in  the  degree  of  </)(a')  and  vice  versa. 

Example.     By  the  method  just  explained  show  that  the  equations 
x2  +  3x  +  2  =  0  and  x  +  1  =  0  have  a  common  root  and  find  this  root. 
.13  2 
Here  D=    110   =1  +  2  —  3  =  0,  so  that  there  is  a  common  root. 

0  1  1 
The  values  of  the  cofactors  of  1  and  3  in  the  first  row  of  D  are  1 
and  —  1.     Hence  the  common  root  is  1  :  —  1,  that  is,  —  1. 


514  A   COLLEGE   ALGEBRA 

927  By  the  preceding  method  either  of  the  unknown  letters 
X,  y  may  be  eliminated  from  a  pair  of  algebraic  equations  of 
the  form  f{x,  y)  =  0,  <^  (a-,  y)  —  0. 

Example.     Solve  x^  -  2  y^  _  x  =  0,  (1) 

2  x2  -  5  ?/2  +  3  ?/  =  0.  (2) 

We  may  regard  (1)  and  (2)  as  quadratics  in  x,  (1)  with  the  coefBcients 
1,  -  1,  -  2  2/2^  and  (2)  with  the  coefficients  2,  0,  -hy-  ^Zy. 
Hence  the  result  of  eliminating  x  is 


1-1        -  2  y2  0 

0        1  - 1  -  2  2/2 

2       0  -  5  ?/2  +  3  ?/  0 

0       2  0  -5  2/2  +  3?/ 


(3) 


which  when  expanded  and  simplified  gives 

2/4  _  (5  y3  _  y2  +  6  2/  =  0.  .  (4) 

Solving  (4),  we  obtain  y  =  0,  1,  -  1,  6. 

It  follows  from  §  924  that  when  y  has  any  of  these  values,  (1)  and  (2) 
are  satisfied  by  the  same  value  of  x.  And  in  fact  when  ?/  =  0,  (1)  and 
(2)  become  x2  —  x  =  0,  2  x^  =  0,  which  have  the  common  root  x  =  0  ;  when 
?/  =  1,  (1)  and  (2)  become  x^  —  x  —  2  =  0,  x^  —  1  =  0,  which  have  the 
common  root  x  =  —  1  since  x^  —  x  —  2  and  x2  —  1  have  the  common  factor 
X  +  1,  §  853 ;  and  so  on.  We  thus  find  that  the  solutions  of  (1)  and  (2) 
are  x,  ?/  =  0,  0  ;  -  1,  1  ;  2,  —  1  ;  9,  6.  The  values  of  x  may  also  be  found 
from  those  of  y  by  applying  §  925  (compare  §  926,  Ex.). 

This  example  illustrates  the  fact  that  if  the  result  of  elimi- 
nating X  from  fix,  y)=  0  and  (f>  (x,  y)=  0  is  R  (y)  =  0,  and  one 
of  the  roots  of  R(y)=  0  is  )8,  the  corresponding  value  or  values 
of  x  can*  always  be  obtained  by  finding  the  highest  common 
factor  of  f(x,  ft)  and  <^  (x,  /3).  Usually  this  highest  common 
factor  will  be  of  the  first  degree,  when  but  one  value  of  x  will 
correspond  to  y  =  /3.  But  it  may  be  of  a  higher  degree,  and 
then  more  than  one  value  of  x  will  correspond  to  y  =  /3. 
928  Properties  of  the  resultant.  Suppose  a  pair  of  equations  to  be 
given  of  the  form 

/(x)  =  a;'"  +  .  • .  +  «,„  =  0,  </.  (x)  =  x"  +  •  •  •  +  i„  =  0. 


DETERMINANTS   AND   ELIMINATION  515 

Let  ai  denote  any  root  of  f{x)  =  0  and  fi^  any  root  of 
^(x)  =  0.  There  will  be  mn  differences  of  the  form  a^  —  /S^; 
let  n  (a^  —  p,?j  denote  their  product. 

Evidently  n  (or,-  —  ^^  =  0  is  the  necessary  and  sufficient  con- 
dition that  one  of  the  roots  a^  be  equal  to  one  of  the  roots  /J^.. 
Moreover,  since  Ti.{cx^  —  ^^.)  is  a  symmetric  integral  function  of 
the  roots  a^  and  of  the  roots  ^^,  it  is  a  rational  integral  func- 
tion of  the  coefficients  of  f{x)  =  0  and  <^(a;)  =  0,  §§  867,  868. 
Hence,  if  B  (/,  </>)  denote  the  resultant  of  f(x)  =  0  and 
(/,  (x)  =  0,  §  923,  we  have 

The  product  n  (a-  —  {3^)  may  be  written 

K-A)('^i-A)---K-)8„), 

(«.-A)(«2-/3.)---(a.-^„), 


K-/3i)K-A)---K-A0- 

But  since  <^ (x)  =  (x  —  (3i) (x  —  /S-.) ■■■(x  —  y8„),  the  product 
of  the  factors  in  the  first  row  is  </>(tri),  in  the  second  row  <f>(a2), 
and  so  on.     Hence 

n(cr, -/?,)  =  <^W-<^(«.)-- -^^K)- 

Again,  since  f(^x)  =  (x  —  a^) (x  —  a,,) ■  ■  -(x  —  a-„,),  the  product 
of  the  factors  in  the  first  column  is  (—  l)"'/(/8i),  in  the  second 
column  (—  l)"'/(/32),  and  so  on.     Hence 

n (a,  -  A.) = (-  lyvm  -/m  ■  ■  -/m- 

When  the  given  equations  have  the  form 

f{x)  =  a.x"'  +  •  •  •  +  a,„  =  0,  <i>  (x)  =  b^x"  +  ...  +  &„  =  0, 

that  is,  when  the  leading  coefficients  are  not  1,  the  product  of 
the  factors  in  the  first  row  is  <l>{a{) /b^,  and  so  on;  and  the 
product  of  the  factors  in  the  first  column  is  (—  l)"'/(/8i)/ao) 
and  so  on.     Hence,  in  this  case,  to  make  11  (a^-  —  yS^)  an  integral 


516  A   COLLEGE    ALGEBRA 

function  of  the  coefficients  of  f(x)  —  0  and  (f>(x)=  0  we  must 
multiply  it  by  ao^'".     We  then  have 

i?(/,  <^)  =  <^™n(a,-;8,) 

=  a'o<l>  (a-i)  •  </>  (0-2)  •  •  •  <^  (or,,.) 

929  In  the  resultant  of  a  pair  of  equations  f  (x)=  0,  ^(x)=  0 
the  coefficients  of  f  (x)  =  0  e7iter  in  the  decree  of  <^(x)  =  0,  and 
vice  versa. 

For  the  product  <j>  (a^)  (^  (a^)  •  ••  <f>  («„)  contains  m  factors, 
each  involving  the  coefficients  of  </>  (a;)  =  0  to  the  first  degree ; 
and  the  product  /(/3i)  -/(ySo)  ■  •  •  /(/8„)  contains  n  factors,  each 
involving  the  coefficients  oif(x)  =  0  to  the  first  degree. 

We  thus  have  another  proof  that  the  determinant  D  described 
in  §§  924,  926  is  the  resultant  of  f(x)  =  0  and  cf>(x)  =  0,  that 
is,  that  D^  R(f,  <f>). 

930  The  su 7)1  of  the  subscripts  of  the  coefficients  of  f(x)  =  0  and 
cf)  (x)  =  0  171  each  term  of  R  (f ,  <f>)  is  van. 

For,  by  §  812,  if  we  multiply  each  coefficient  of  f(x)  and 
<^  (x)  by  the  power  of  r  indicated  by  its  subscript,  we  obtain 
two  equations 

/i  (x)  =  aoX"*  +  miCC™-  ^  +  r^aox'"-  ^ -\ \-  r"'a„  =  0, 

<^i  (x)  —  boX"  +  j'bix"  ~  ^  +  7'%„x" '-  +  •  •  •  +  r''b,^  =  0, 

whose  roots  are  r  times  the  roots  oi  f(x)  =  0  and  ^(.t)  =  0. 

Each  term  of  R  (f,  <f>i)  will  be  equal  to  the  corresponding 
term  of  R  (f  (f>)  multiplied  by  a  power  of  r  whose  exponent  is 
the  sum  of  the  subscripts  of  the  coefficients  of /(a-)  and  <f>(x) 
which  occur  in  the  term.  Hence  our  theorem  is  demonstrated 
if  we  can  show  that  in  every  term  this  exponent  is  m?i.  But 
since  there  are  m?i  factors  in  the  product  n  (a,  —  yS^.),  we  have 

R  (./;,  <^0  =  a'l/j'i;  n  (/-a,  -  7'/3,)  =  7"""  ■  R  (f  <^). 


detp:rminants  and  elimination         517 

Discriminants.    The  discriminant  oif(x)  —  a^"^  -\ 1-  «„  =  0     931 

is  that  integral  function  of  the  coefficients  oi  fix)  whose  vanish- 
ing is  the  necessary  and  sufficient  condition  that/(a:-)  =  0  have 
a  multiple  root  (compare  §§  635,  873). 

If  D  denote  the  discriminant  of  /(*")  =  0,  then 

For,  by  §  851,  f{x)  =  0  has  a  finite  multiple  root  when  and 
only  when  f{x)  —  0  and  /'  (cc)  =  0  have  a  finite  root  in  common. 
By  §  928,  the  condition  that/(.r)  =  0  and  /'  {x)  =  0  have  a  root 
in  common  is  R(^f,f')=  0.  But  Uq  is  a  factor  of  R (/,/'),  as 
may  be  shown  by  expressing  R  (/,  /')  in  the  determinant  form 
of  §924.  Hence  R(f,f')=0  when  «« =  0.  But  the  root 
which  in  this  case  is  common  to  f(x)  =  0  and/' (x)  =  0  is  oo, 
§81G.  Tlierefore,  since  oo  is  not  a  multiple  root  of /(a;)=  0 
unless  both  a^  and  a^  vanish,  we  have  D  —  R(f,f')/aQ. 

Thus,  for  aox2  +  «iX  +  a2  =  0,  Z>  = 

The  diserimijiant  of  f  (x)  =  0  is  equal  to  the  product  of  the     932 

squares  of  the  differences  of  every  t%vo  of  the  roots  of  f  (x)  =  0 
multiplied  by  a  certain  power  of  the  leading  coefficient  ao. 

Thus,  if  /(x)  =  ao  (X  -  ft)  (X  -  ft)  (x  -  ft),  (1) 
then,  §  865,            f'{x)  =  a^  [(x  -  ft)  (x  -  ft)  +  (x  -  ft)  (x  -  ^i) 

+  (x-ft)(x-ft)].  (2) 

By  §  928,      B  (/,  /')  =  alf  (^i)/'  (ft)/'  (ft).  (3) 

But  from  (2),   /'(/3i)  =  «o(/3i  -  ft)  (/3i  -  ft),  and  so  on.  (4) 

Substituting  (4)  in  {?,)  and  simplifying,  we  have 

R  if,  n  =  -  aS  (^1  -  ^2)-  (ft  -  ft)-  (ft  -  ft)2, 
whence  B  =  -  a^ifi^  -  ft)--2 (ft  -  ft)2 (ft  -  ft)2.  (5) 

On  the  number  of  solutions  of  a  pair  of  equations  in  two  unknown     933 
letters.     Observe  that  if  in  the  equation  f(x,  y)=  0  we  make 
the  substitutions  x  =  Xi/x^,  y  =  x^/x^,  and  clear  of  fractions, 
we  transform  f{x,  y)  =  0  into  a  homogeneous  equation  of  the 


flo         dl        ^2 

2ao   ai     0 

-^  ao  = 

-  (a^  -  4  aoai). 

0        2ao  ai 

518  A   COLLEGE   ALGEBRA 

same  degree  in  Wi,  a-g,  x^.     Thus,  x^  +  xy  +  y-{-l  =  0  becomes 

*t/i   ~|~  *)C-^pCt^^^  "Y"  OC^'iCo  ~j~  OCq  —  U. 

Observe  also  that  a  homogeneous  equation  of  the  wth  degree 
in  iCg,  Xg  which  is  not  divisible  by  Xg  determines  n  finite  values 
of  the  ratio  Xo  /x^.     Thus,  from  a-|  —  3  x^x^  +  2x1  — Owe  obtain 
x^/x^  =  1  or  2. 
934        Let  f(x,  y)  =  0  and  <^  (x,  y)=0 

denote  two  equations  whose  degrees  are  ?n  and  ?i  respectively. 
If  they  involve  the  terms  a;"*  and  x",  then  by  substituting 
x  —  Xi/xs,  y  =  x^lxz,  clearing  of  fractions,  and  collecting 
terms,  we  can  reduce  them  to  the  form 

F(a-i,  a-2,  x^  =  ao^cr  +  ai^^  "  ^  H h  «,„  =  0,  (1) 

$(a:i,  a-2,  x^  =  b^xl  +  h^xl'^  -[ \-  b,^  =0,  (2) 

where  each  of  the  coefficients  ciq,  a^,  ■■■,  bo,  b^,  •••  denotes  a 
homogeneous  fmiction  of  Xo,  x^  of  the  degree  indicated  by  its 
subscript.  Hence  R,  the  resultant  of  (1)  and  (2)  with  respect  to 
Xi,  is  a  homogeneous  function  of  X2,  x^  of  the  degree  ?»n,  §  930. 
By  §  928  the  necessary  and  sufficient  condition  that  (1)  and 
(2)  be  satisfied  by  the  same  value  of  x^  is  that 

^  =  0.  (3) 

If  R  is  not  divisible  by  X3,  then  R  =  0  is  satisfied  by  ??m 
finite  values  of  x^/x^  or  y,  §  933.  If  /?  denote  any  one  of 
these  values,  the  equations  f(x,  (3)=  0,  <j>  (x,  /3)=  0  have  a 
common  root,  and  if  this  root  be  a,  then  x  =  a,  y  =  fi  is  a 
solution  oif(x,  y)  =  0,  <f,  (x,  y)  =  0  (compare  §  927).  Moreover 
it  can  be  shown  that  to  each  simple  root  oi  R  =  0  there  thus 
corresponds  a  single  solution  of  f(x,  y)  =  0,  cf}  (x,  y)  =  0  ;  and 
that  to  a  multiple  root  of  order  r  oi  R  =  0  there  correspond 
r  solutions  of  f(x,  y)=  0,  <j>(x,  y)  =  0,  all  different  or  some 
of  them  equal.  Hence  f(x,  y)  —  0,  <f)  (x,  y)—0  have  mn  finite 
solutions. 

If  R  is  divisible  by  x^,  then  72  =  0  is  satisfied  by  only 
mn  —  fi.  finite  values  of  x^/x^  or  y,  and  therefore  f(x,  y)  =  0, 


DETERMINANTS    AND   ELIMINATION  519 

<^  (^x,  y)  =  0  have  only  mn  —  fx  finite  solutions.  But  since 
X  =  Xi/x^  and  y  =  x^/x^,  when  iCg  =  0  either  x  ov  y  or  both 
X  and  y  are  infinite.  We  therefore  say  in  this  case  that 
f{x,  y)=  0,  (f)  (x,  y)  =  0  have  /x  infinite  solutions. 

If  the  given  equations  f(x,  y)=  0,  (ft  (x,  y)—0  lack  the  x"* 
and  x"  terms,  we  can  transform  them,  by  a  substitution  of  the 
form  y  =  y'  -\-  ex,  into  equations  of  the  same  degrees  which  have 
these  terms.  By  what  has  just  been  proved  the  transformed 
equations  in  x,  y'  will  have  mn  solutions.  But  if  x  =  a,  y'  =  /3 
be  any  one  of  these  solutions,  then  x  =  a,  y  =  (3  +  ca  is  &  solu- 
tion otf(x,  tj)  =  0,ct>  (x,  y)  =  0.  Hence /(a-,  y)  -  0,  <^  {x,  y)  =  0 
also  have  mn  solutions. 

In  the  preceding  discussion  it  is  assumed  that  R  does  not 
vanish  identically.  If  R  does  thus  vanish,  f(x,  y)  and  <^(x,  y) 
have  a  common  factor  and  therefore  f{x,  ?/)  =  0,  ^  {x,  y)  =  0 
have  infinitely  many  solutions. 

We  therefore  have  the  theorem  : 

If  i(x,  y)  and  4>(x,  y)  are  of  the  degrees  m  and  n  respec- 
tively and  have  no  common  factor,  the  equations  f(x,  y)=0, 
<^  {x,  y)—0  have  mn  solutions. 

EXERCISE  LXXXV 

1.  By  the  method  of  §§  924,  925  show  that  the  equations  6  x2+5  x-6  =  0 
and  2x3  +  x2  —  9x  —  9  =  0  i^ave  a  common  root  and  find  this  root. 

2.  Form  the  resultant  of  aoX-  +  aiX  +  a2  =  0  and  hoz-  +  6iX  +  60  =  0. 

3.  Find  the  resultant  of  ax^  +  hxr  +  ex  +  fZ  =  0  and  x^  =  1. 

4.  By  the  method  of  §  931  find  the  discriminants  of  the  equations 
(1)  x3  +  px  +  g  =  0.  (2)  ax3  +  &x2  +  c  =  0. 

5.  By  aid  of  §  931  show  that  x^  +  x^  -  8  x  -  12  -  0  has  a  double  root 
and  find  this  root. 

6.  Solve  the  following  pair  of  equations  by  the  method  of  §  927. 

x2  -  3  x?/  +  2  2/2  -  16  X  -  28  y  =  0, 
x2  —  xy  —  2  2/2  _  5  X  —  5  2/  =  0. 


520  A    COLLEGE    ALGEBRA 

XXXII.     CONVERGENCE    OF   INFINITE   SERIES 

DEFINITION   OF   CONVERGENCE 

935  Infinite  series.     If  i/^,  v. 2,  •  •  •,  ic^,  •  •  ■  denotes  any  given  never- 
ending  sequence  of  numbers,  §  187,  the  expression 

"1  +  ?'2  H \-  iin  -\ 

is  called  an  infinite  series  (compare  §  704). 

For  Vi  +  n.  +  ■■■  we  may  write  2»„,  read  " sum  of  n„  to 
infinity." 

The  series  2«„  is  called  real  when  all  its  terms  Vi,  «„,  •  •  •  are 
real,  jjositlve  when  all  its  terms  are  positive.  In  what  follows 
we  shall  confine  ourselves  to  real  series. 

A  series  is  often  given  by  means  of  aformula  for  its  nth  term  m„. 
Thus,  if  M„  =  Vn/{n  +  1),  the  series  is  Vl/2  +  V2/3  +  V3/4  -\ . 

Sometimes  such  a  formula  is  indicated  by  writing  the  first  three  or  four 

terms  of  the  series.     Thus,  in  1/2  +  1  •  3/2  •  4  +  1  •  3  •  5/2  •  4  •  6  H we 

have  u„  =  1  ■  3  •  ■  •  (2  u  -  l)/2  •  4  •  •  •  2  ji. 

936  Convergence  and  divergence.     Let  S„  denote  the  sum  of  the  first 

71  terms  of  the  series  u^  +  v.,  +  •  •  •,  so  that  Sj  =  Ui,  So  =  Ui  +  u^, 
and  in  general  S„  =  ir^  -\-  v„  -\-  ■  ■ .  -}-  y^. 

As  n  increases,  *S',^  will  take  successively  the  values  ?<i, 
Ml  +  Wjj  Ml  +  ^2  +  «3,  •  •  •,  and  one  of  the  following  cases  must 
present  itself,  namely  : 

S„  will  approach  some  finite  number  as  limit, 
or  S„  will  approach  infinity, 

or  .§„  will  be  indeterminate. 

In  the  first  case  the  series  Uy  +  Uo  +  •  ■■  is  said  to  be  conver- 
gent, and  lim  .V„  is  called  its  sum.  In  the  second  and  third 
cases  the  series  is  said  to  be  divergent. 

When  Ml  +  M2  +  •  •  •  is  convergent  we  may  represent  its  sum, 
lim  S,^,  by  S  and  write   5  =  Mi  +  Mj  -| ,  that   is,  we  may 


CONVERGENCE   OF   INFINITE    SERIES  521 

regard  the  series  as  merely  another  expression  for  the  definite 
number  S. 

Thus,  the  geometric  series  1/2  +  1/4  +  1/8  +  1/16  +  •  •  •  is  convergent 
and  its  sum  is  1.  For  here,  as  n  increases  /S„  takes  successively  the  values 
1/2,  3/4,  7/8,  15/16,  •  •  •  and,  as  is  proved  in  §  704,  it  approaches  1  as 
limit.     Observe  that  here,  as  in  every  convergent  series,  lim  w„  =  0. 

The  series  1  +  1  +  1  +  •  •  •  is  divergent.  For  jS„  takes  successively  the 
values  1,  2,  3,  •  •  •  and  therefore  approaches  go. 

The  series  1  —  1  +  1  —  l  +  ---is  divergent.  For  S„  takes  successively 
the  values  1,  0,  1,  0,  •  •  •.     It  is  therefore  indeterminate. 

We  therefore  have  the  following  definitions  :  937 

An  infinite  series  is  said  to  be  convergent  when  the  sum  of 
its  first  n  terms  approaches  a  finite  limit  as  n  is  indefinitely 
increased.      Otherwise  it  is  said  to  he  divergent. 

The  limit  of  the  sum  of  the  first  n  terms  of  a  convergent 
series  is  called  the  sum  of  the  series. 

This  is  a  new  use  of  the  word  sum.  Hitherto  sum  has  meant  the 
result  of  a  finite  number  of  additions  performed  consecutively;  here  it 
means  the  limit  of  such  a  result.  It  must  therefore  not  be  assumed  that 
the  characteristic  properties  of  finite  sums,  namely,  conformity  to  the 
commutative  and  associative  laws,  always  belong  to  these  infinite  sums 
(see  §§  941,  961). 

In  determining  whether  a  given  series  is  convergent  or  diver-    938 
gent,  a  finite  number  of  its  terms  may  be  neglected. 

For  the  sum  of  the  neglected  terms  will  have  a  definite  finite  value. 

If  tiy  +  «2  4-  •  •  •  (1)  is  a  convergent  series  having  the  sum  S,     939 
and  c  is  any  finite  number,  then  cu^  +  cu^  +  •  •  •  (2)  is  a  con- 
vergent series  having  the  sum  cS.     But  if  (1)  is  divergent,  so 
is  (2). 

For  if  the  sum  of  the  first  n  terms  of  (1)  be  iS„,  the  sura  of  the  first 
71  terms  of  (2)  is  cS„ ;  and  lim  cSn  —  c  lim  S„  =  cS. 

The  sum  of  a  convergent  series  will  not  be  changed  if  its     940 
terms  are  combined  in  groups  without  changing  their  order. 


622  A   COLLEGE   ALGEBRA 

Thus,  if  the  given  series  be  ui  +  «2  H >  and  grj,  ^2,  •  •  •  denote  the  sums 

of  its  first  two  terms,  its  next  four  terms,  and  so  on,  the  series  fifi  +  g'2  H 

will  have  the  same  sum  as  Wi  +  t<2  +  ■  •  •  • 

For  if  Un  denote  the  last  term  in  the  group  g,n,  we  have 

gi  +  92  +  ■■■  +  9m  =  Ul  +  Mo  +  ...  +  «„, 

and  the  two  members  of  this  equation  approach  the  same  limit  as  m  and 
therefore  n  is  indefinitely  increased. 

In  the  same  manner  it  may  be  shown  that  a  divergent  positive  series 
remains  divergent  when  its  terms  are  grouped. 

941  We  may  therefore  introduce  parentheses  at  will  in  a  con- 
vergent series.  It  is  also  allowable  to  remove  them  unless,  as 
in  the  following  example,  the  resulting  series  is  divergent. 

The  convergent  series  1/2  +  1/4  +  1/8  +  •  •  •,  §  936,  may  be  written 
(11  _  1)  _f  (1|  _  1)  +  (11  _  1) -I- . .  ..  But  here  it  is  not  allowable  to 
remove  parentheses  since  IJ  -  1  +  1^  —  1  +  1^  —  1  +  •  •  •  is  divergent. 

942  It  is  sometimes  possible  to  find  the  sum  of  a  series  by  the 
removal  of  parentheses. 

Thus,  the  sum  of  1/1  •  2  +  1/2  •  3  +  1/3  •  4  +  ■  •  •  is  1. 

For  Sn  =  —  +  —  +  •  •  ■  + 

1-2      2-3  ?i(?i  +  l) 

1223  n      n+1  n+1 

Hence         -S  =  liin  S„  =  lim  (l ")  =  1. 

V         n  +  1  / 

Example.     Find  the  sum  of  the  series  whose  nth  term  m„  is  1  /?i  (n  +  2). 

943  Remainder  after  n  terms.     If  the  series  «i  +  t'2  H (1)  is 

convergent,  that  portion  of  the  series  which  follows  the  nth. 
term,  namely,  ??„  +  ,  +  ?^,  +  ._,  +  •••  (2),  will  also  be  convergent, 
§  938.  Let  R„  denote  the  sum  of  (2).  It  is  called  the 
remainder  after  n   terms  of  (1). 

Evidently  lim  /.'„  =  0. 


CONVERGENCE   OF   INFINITE   SERIES  523 


POSITIVE  SERIES 

Theorem  1.    A  positive  series  n^  +  «2  +  •  •  •  *«  convergent  if,  as     944 
n  increases,  S^  remains  always  less  than  some  finite  number  c. 

For  since  the  series  is  positive,  .S",^  continually  increases  as 
n  increases.  But  it  remains  less  than  c.  Hence,  §  192,  it 
approaches  a  limit.    Therefore,  §  937,  the  series  is  convergent. 

Theorem  2.     Let  Ui  +  Uj  H (1)  denote  a  given  positive  series,     945 

a7id  let  a^  -j-  c''2  +  ■  ■  ■  (2)  denote  a  positive  series  known  to  he 
convergent.     The  series '(1)  is  convergent  in  any  of  the  cases: 

1.  When  each  term  of  (1)  is  less  than  the  corresponding  term 
of  {2). 

2.  When  the  ratio  of  each  term  of  (1)  to  the  correspc^nding 
■term  of  (2)  is  less  than  some  finite  number/ a.  ) 

3.  When  in  (1)  tlie  ratio  of  each  term  ^  the  immediately 
preceding  term   is  less  than  the  corresponding  ratio  in  (2). 

1.  For  let   5„  denote    the    sum    of   the    first    n   terms    of 

"i  +  "2  H J  and  let  A  denote  the  sum  of  the  series  a^  +  ag  -I • 

If  Ui  <  «!,  i<2  <  ^2)  •  ■  ■)  we  shall  always  have  S^  <  A.     Hence 
«i  +  «2  +  •  •  ■  is  convergent,  §  944. 

2.  For  if         —  <  c,  —  <  c,  ••■,  then  u-^  <  ca^  ««  <  caz,  ••■. 

Ol  «2 

Therefore,  since  cai-{-ca2-] is  convergent,  §  939,  the  series 

III  +  ^2  +  •  •  •  is  convergent, by  1. 


3. 

For  if 

Ui          Oi 

«2           «2 

«4           04 

Us       as 

en 

W2  ^  «i 
—  <  — > 

rt2           «1 

«3           «2 

W4           «3 

a4       a3 

It  follows  from  these  inequalities  that  each  of  the  ratios 
Uo/a^,  Us/ds,  •  •  •  is  less  than  the  finite  number  Wi/^i.  There- 
fore «i  +  M2  +  •  •  •  is  convergent,  by  2. 


524  A   COLLEGE    ALGEBRA 

It  follows  from  §  938  that  the  same  conclusions  can  be 
drawn  if  any  one  of  the  relations  1,  2,  3  holds  good  for  all  but 
a  finite  number  of  the  terms  of  the  series  (1)  and  (2). 

Example.     Prove  that 

1  +  1/2  +  1/2  •  3  +  1/2  •  3  •  4  +  •  ■ .  (1) 

is  convergent  by  comparing  it  witli  the  convergent  geometric  series 

1  +  1/2  +  1/2  •  2  +  1/2  .  2  •  2  +  . . .  (2) 

by  each  of  the  methods  1,  2,  3. 

First.  Each  term  of  (1)  after  the  second  is  less  than  the  corresponding 
term  of  (2).     Hence  (1)  is  convergent,  by  1. 

Second.  The  ratios  of  the  terms  of  (1)  to  the  corresponding  terms  of  (2), 
namely,  1,  1,  2/3,  2  •  2/3  •  4,  •  ■  • ,  are  finite.    Hence  (1)  is  convergent,  by  2. 

Third.  The  ratios  of  the  terms  of  (1)  to  the  immediately  preceding 
terms,  namely,  1/2,  1/3,  1/4,  •  •  • ,  are  less  than  the  corresponding  ratios 
in  (2),  namely,  1/2,  1/2,  1/2,  •  •  •.     Hence  (1)  is  convergent,  by  3. 

946  Theorem  3.    Let  Ui  +  Uo  -| (1)  denote  a  given  positive  series, 

and  let  bj  +  bj  +  •  •  •  (2)  denote  a  positive  series  known  to  he 
divergent.     The  series  (1)  is  divergent  in  any  of  the  cases: 

1.  Wheyi  each  term  of  (1)  is  greater  than  the  corresponding 
term  of  (2). 

2.  When  the  ratio  of  each  term  of  (1)  to  the  corresponding 
term  of  (2)  is  greater  than  some  positive  number  c. 

3.  When  in  (1)  the  ratio  of  each  term  to  the  immediately 
preceding  term  is  greater  than  the  corresponding  ratio  in  (2). 

The  proof  of  this  theorem,  which  is  similar  to  that  given  in 
§  945,  is  left  to  the  student. 

947  Test  series.  The  practical  usefulness  of  the  preceding  tests, 
§§  945,  946,  evidently  depends  on  our  possessing  test  series 
known  to  be  convergent  or  divergent.  The  most  important 
of  these  test  series  is  the  geometric  series  a  +  ar  -\-  ar^  -\-  ■  ■ -, 
which  has  been  shown,  §  704,  to  be  convergent  when  r  <  1, 
and  which  is  obviously  divergent  when  r>l.  Another  very 
serviceable  test  series  is  the  following. 

948  The  series  1  +  1/2p  +  1/3p  +  •  ■  •  +  I/up  +  •  •  •  , 
when  p  >  1,  divergent  when  p  <  1. 


CONVERGENCE    OF    INFINITE    SERIES  525 

1.  p  >  1.  Combining  the  two  terms  beginning  with  1/2^, 
the  four  terms  beginning  with  1/4^,  the  eight  terms  beginning 
with  1/8^,  and  so  on,  we  obtain  the  equivalent  series,  §  940, 


^<h-hHh-h-h-hy 


(1) 


Evidently  each  term  of  (1)  after  the  first  is  less  than  the 
corresponding  term  of  the  series 

that  is,  less  than  the  corresponding  term  of 

^  +  l  +  i  +  ---^orl+^^  +  -^_~  +  :..       (3) 

But  since  p  >  1,  and  therefore  1/2^"'  <  1,  the  geometric 
series  (3)  is  convergent.     Hence  (1)  is  convergent,  §  945,  1. 

2.  J)  —  1.  Combining  the  two  terms  ending  with  1/4,  the 
four  terms  ending  with  1/8,  the  eight  terms  ending  with 
1/16,  and  so  on,  we  obtain 


-^(^0-G-^^O 


(4) 


Evidently  each  term  of  (4)  after  the  second  is  greater  than 
the  corresponding  term  of  the  series 

that  is,  greater  than  the  corresponding  term  of 

12      4  111 

1+2  +  4  +  8  +  -,  orl  +  2  +  2  +  2  +  -.  («) 

But  (6)  is  divergent.     Therefore  (4)  is  divergent,  §  946,  1. 

3.  p  <1.     In   this  case  the   series  1  +  1/2"  +  1/3^  -] 

is  divergent  since  its  terms  are  greater  than  the  corresponding 
terms  of  the  series  1  +  1/2  +  1/3  +  •  •  •,  which  has  just  been 
proved  to  be  divergent,  §  946,  1. 


526  A    COLLEGE    ALGEBRA 

949  Applications  of  the  preceding  theorems.  The  following  exam- 
ples will  serve  to  illustrate  the  usefulness  of  the  theorems  of 
§§  945,  946. 

Example  1.     Show  that  1/1  •  2  +  1/2  •  3  +  1/3  •  4  +  ■  •  •  is  convergent. 

It  is  convergent  because  its  terms  after  the  first  are  less  than  the 
corresponding  terms  of  the  convergent  series  1/2^  +  I/32  +  I/42  +  • .  • , 
§945,  1. 

Example  2.     Show  that  1  +  1/3  +  1/5  +  1/7  +  •  •  •  is  divergent. 

The  ratios  of  the  terms  of  this  series  to  the  corresponding  terms  of  the 
divergent  series  1  +  1/2  +  1/3  +  1/4  +  •  •  • ,  namely,  1,  2/3,  3/5,  4/7,  •  •  • , 
71/(2  n  —  1),  are  all  greater  than  1/2.  Hence  1  +  1/3  +  1/5  +  •  •  •  is 
divergent,  §  946,  2. 

Example  3.  Is  the  series  In  which  m„  =  {2n  +  l)/{n^  +  n)  convergent 
or  is  it  divergent  ? 

Here  ^^^^_2n  +  l_  n    2  +  1/n  ^1     2  +  1/n 

v?  +  n      n^    1  +  1/n^      n^    1  +  1/n^' 

Hence  the  ratio  of  m„  to  1/n^  is  (2  +  l/n)/(l  +  l/n^),  an  expression 
which  is  finite  for  all  values  of  n,  and  which  approaches  the  finite  limit 
2  as  n  increases.  But  1/n^  is  the  nth  term  of  the  convergent  series 
1  +  1/2-  +  1/3-  +  •  ■  • .    Therefore  the  given  series  is  convergent,  §  945,  2. 

950  By  the  method  employed  in  Ex.  3,  it  may  be  proved  that  if 
7i„  has  the  form  n,^  =  f(7i)/<l>(n),  where  f(n)  and  <^(??)  denote 
integral  functions  of  71,  the  series  is  convergent  when  the 
degree  of  <^  (n)  exceeds  that  of /(?i)  by  more  than  1 ;  otherwise, 
that  it  is  divergent. 

Example  1.     Show  that  the  following  series  are  convergent. 


1  1 


> 


a(a  +  b)      (a  +  b){a  +  2b)'^  {a  +  2b){a  +  3b)'^"'' 
Example  2.     Show  that  the  following  series  are  divergent. 

V2       V3       V4  1-1-2  V2      1  +  3  V3      I+4V4 


CONVERGENCE    OF   INFINITE    SERIES  527 

Example  3.  Write  out  the  first  four  terms  of  the  series  in  which  m„ 
has  each  of  the  following  values  and  determine  which  of  these  series  are 
convergent  and  which  divergent. 

2n-l  .^,  V^  ,„,  n2_(^_l)2 

(1)    Un  = (2)    Un  = (3)    Un  —  ^^ -• 

^  '  (n  +  1)  (n  +  2)  n2  +  1  n^  +  (n  +  1)3 

Theorem  4.     The  positive  series  u^  +  u^  +  ■  •  •  is  convergent  if     951 
the  ratio  of  each  of  its  terms  to  the  immediately  preceding  term 
is  less  than  some  number  r  which  itself  is  less  than  1. 

For  in  «i  +  n^  +  1/3  +  •  •  •  (1)  the  ratio  of  eacli  term  to  the 
immediately  preceding  term  is  less  than  the  corresponding 
ratio  in  the  geometric  series  Vi  +  Uir  +  Uir"^  +  •  •  •  (2),  since  in 
(1)  the  ratio  in  question  is  always  less  than  r,  while  in  (2)  it 
is  equal  to  7\  But  (2)  is  convergent  since  r  <  1.  Therefore 
(1)  is  convergent,  by  §  945,  3. 

If  the  ratios  above  mentioned  are  equal  to  1  or  greater  than 
1,  the  series  is  divergent ;  for  in  this  case  lim  «„  ^  0. 

Corollary.    If  as  n  increases  the  ratio  u„  +  i/Uj,  apjjroaches  a     953 
definite  limit  X,  the  series  is  convergent  ichen  A  <  1,  divergent 
when  A  >  1. 

1.  For  if  X  <  1,  take  any  number  r  such  that  X  <  r  <  1. 
Then,   since   lim  (?/,„  + j/^^,)  =  X,  after  a  certain  value  of  n 

we  shall  always  have  Vn  +  i/Un  —  X  <  r  —  X,  §  189,  and  there- 
fore u„  +  i/i(„  <  r.    Hence  the  series  is  convergent,  §§  938,  951. 

2.  If  X  >  1,  after  a  certain  value  of  71  we  shall  always  have 
^^<  +  i/"n  >  1-     Hence  the  series  is  divergent,  §  951. 

When  «„  +  i/«„  >  1  and  lim  (m„+i/?<„)  =  1,  the  series  is 
divergent;  but  when  «„  +  i/'/„  <  1  and  lim (w„_^i /?/„)  =  1,  no 
conclusion  can  be  drawn  from  the  theorem  of  §  951. 

Example  1.     Show  that  -  +  ——-  +       "^"    ^  -f  •  ■  ■  is  convergent. 
5       5  •  10      5  •  10  •  15 

The  ?xth  term  of  this  series  is  3  •  5  •  7  •  •  •  (2  n  +  l)/5  ■  10  •  15  •  •  •  5  n,  and 
the  ratio  of  this  term  to  the  term  which  precedes  it  is  (2  n  +  l)/5  n. 

But  since  (2  n  +  l)/5  n  =  2/5  +  1/5  n,  lim  (2  n  +  l)/5  n  =  2/5,  which 
is  <  1.     Hence  the  series  is  convergent. 


528  A   COLLEGE    ALGEBRA 


Example  2.     When  is  ■ 1 1 f-  •  •  •  convergent, 

X  being  positive?  1  +  x       1  +  2x^       1  + 3x3 

M„  +  i  _  I  +  nx"  X"  +  l/n 


Here 


l  +  (>i  +  l)x"  +  i      x»  +  i(l  +  l/n)  +  l/ft 


M  1 

and  therefore  lim  -^^^  —  -  • 

Un  X 

Hence  the  series  is  convergent  when  1/x  <  1,  that  is,  when  x  >1. 

1       1  •  3      1  •  3  •  5 
Example  3.     Show  that  -  -\ '-  ^ '^^ 1-  •  •  •  is  convergent. 

1  1-4      1-4-7  ^ 

Example  4.     Show  that  -  +  - 1 1 is  convergent. 

2  2    5      2    5    2 

X      x^      x^ 
Example  5.     When  is  -  H 1 f-  •  •  •  convergent,  x  being  positive  ? 

Example  6.    When  is 1 h 1 convergent,  x  being 

positive?  1+^      1+^'^      1  +  -^' 

953         Series  in  which  lim  (Un  +  i/Un)  =  1.    In  a  series  of  this  kind  the 
ratio  ii„  +  i/Un  can  be  reduced,  to  the  form 

Wn  +  i/»„  =  l/(l  +  «-„M 
where  lim  (a„/n)  —  0.  We  proceed  to  show  that  if,  as  n 
increases,  cr„  ultimately  becomes  and  remains  greater  than 
some  number  which  is  itself  greater  than  1,  the  series  is  con- 
vergent ;  but  that  if  a„  ultimately  becomes  and  remains  less 
than  1,  the  series  is  divergent. 

1.    For  suppose  that  after  a  certain  value  of  n,  which  we  may  call  k,  we 
have  cTb  >  1  +  a,  where  a  is  positive. 

Then         ^!^  =  ;; < ,  when  n%k. 

Un        1  +  a„/n     1  +  (1  +  a)/n 

But  we  may  reduce  this  inequality  to  the  form 

Un  +  i<  —  [nun-(n  +  l)u„  +  i],  when  n%k.  (1) 

In  (l)  set  n  =  k,  k  +  I,  ■  ■  ■ ,  k  +  I  —  1  successively,  and  add  the  resulting 
inequalities.     We  obtain 

Uk  +  i  +  UK-  +  2  +  ■■■  +  ui-  +  i<  ~[kuk  -  (k  +l)u^.+  ,].  (2) 

It  follows  from  (2)  that  as  I  increases  the  sum  of  the  first  I  terms  of 
the  positive  series  Uk  +  i  +  Uk  +  2-^ remains  always  less  than  the  finite 


CONVERGENCE   OF    INFINITE    SERIES  629 

number  ku^/a,  which  proves  that  this  series  is  convergent,  §  944.    There- 
fore the  complete  series  vi  +  M2  +  •  •  •  is  convergent,  §  938. 
2.    Suppose  that  when  n  >  A;  we  have  «»  <  1. 

Then  -^^^  = > ,  when  n>k. 

w„        1  +  a„/n     1  +  1/n 

But  1/(1  +  l/ri)  is  the  ratio  of  the  corresponding  terms  of  the  divergent 
series  1  +  1/2  +  1/3  +  •  •  ■ ;  for  l/(n  +  1)  ^  1/n  =  1/(1  +  1/n). 
Hence  the  given  series  Mi  +  mj  +  •  •  •  is  divergent,  §  946,  3. 

If  a„  remains  greater  than  1  but  approaches  1  as  limit,  the 
preceding  test  will  not  determine  whether  the  series  is  conver- 
gent or  divergent.  But  in  this  case  «"„  can  be  reduced  to  the 
form  a„  =  1  +  (3,,/n,  where  lim  ;8„/?i  =  0 ;  and  if  j3„  remains 
less  than  some  finite  number  b,  the  series  is  divergent. 
For  since  ^„  <  6,  we  have 

Un  +  l  _         1           _                1                                 1 
M„    "  1  +  a„/n  ~  1  +  1/n  +  /3„/n2     l  +  l/n  +  b/n^' 
But  1/(1  +  1/n  +  b/n'^)  in  turn  is  greater  than  the  ratio  of  the  corre- 
sponding terms  of  the  divergent  series  1/(1  — 6) +  1/(2 —  6) +  1/(3  — 6)  H . 

1  1  n-b  1  1 


For 


{n  +  l)-bn-b      (n  -  6)  +  1      l  +  l/(n-6)     l  +  1/n  +  b/n^ 
111  1       6       62 


+  -,  +  -^  + 


n  —  b      n  1  —  b/n      n      n^      n^ 
Therefore  the  given  series  iti  +  W2  +  •  •  •  is  divergent,  §  946,  3. 
It  follows  from  the  preceding  discussion  that  a  series  in     954 
which  Un  +  i/i(n  can  be  reduced  to  the  form 

v,^^Ju,^  =  (iiP  +  ««''-!  H )/(nP  +  a'nP-'^  -\ ) 

is  convergent  when  a'  —  a  >  1,  divergent  when  a'  —  a<_  1. 
For  dividing  the  denominator  of  this  fraction  by  its  numerator,  we  have 
u„^i  _  np  +  anP-^  +  ■■  ■  _  1 

Un    ~  ni>  +  a'nP  - 1  H ~  I  +  (of  -  a)/n  +  ^JrC^' 

where  /3„  is  finite. 

Example.     Prove  that  the  "  hypergeometric  series" 

a-/3      a(a  +  l)^(^  +  l)       a(a  +  1)  (a  +  2)/3(^  +  1)  (^  +  2) 
■^1-7  1-27(7  +  1)  l-2-37(7  +  l)(7  +  2) 

is  convergent  when  7  —  a  —  /3  >  0,  divergent  when  7  —  a  —  )3  ^  0. 


630 


A   COLLEGE    ALGEBRA 


EXERCISE   LXXXVI 

Determine  whether  the  following  series  are  convergent  or  divergent. 
1 


1  1 

2+1  "^22  +  1      23+1 


3. 

-^  +  ^  +  ^H 
2-3      3-4      4-5 

5. 

1           1          1 

V3       ^       ^ 

7. 

2      2-4       2-4.6 
4      4-7      4  •  7  •  10 

8. 

2      2-3      2-3.4 

4      4-5      4.5C 

9. 

1       1  •  3       1  •  3  •  5 

2  '  2-4      2-4-6 

a^      a^  +  1      a'2  +  2 


2  •  4  •  6  •  •  •  2  /I 


4  •  7  •  10  •  •  •  (3  n  +  1) 


2-; 


(n+l) 


+  •••  + 


4  ■  5  •  6  •  •  •  (ji  +  3) 
1.3..5...(2n-l) 


2-4-6---2?i 

Write  out  the  first  four  terms  of  the  series  in  which  ?t„  has  the  follow- 
ing values  and  determine  whether  these  series  are  convergent  or  divergent. 

3/— 

n  +  l  ^,  Vn 


10.  u„  = ■ 11.    u„  

12.     Un  =  V?l-  +  1   —  ?l  =  


'Vn-  +  1  +  ?i 
Determine  wliether  the  series  in  which  itn  +  \/Un  has  the  following 
lues  are  convergent  or  divergent. 

u„  +  i  2n  ,  ^     u„  + 1  _    3  ?i3  —  2  ?i2 

~  3  n^  +  9i2  +  1  ■ 


13. 


14. 


Un        2  n  +  3 

For  what  positive  values  of  x  are  the  following  series  convergent? 

,.     .       3         3-6   ,       3-6-0    , 

15.    1  +  -X  + a;2  + x3  +  .... 

5         5-8  5-8.11 


16. 


1  +  X      1  +  X- 


1  +  X^         1  +  X'* 

a(a  +  1)      a(a  +  l)(a  +  2) 
1-2  1  •2-3 


is    divergent 


17.  Show    that    -  + 
when   a  is  positive. 

18.  If  for  all  values  of  n  we  have  Vm^  <  r,  where  r  is  positive  and  less 
than  1,  show  that  Wi  +  uo  +  •  •  ■  is  convergent  by  comparing  it  with  the 
convergent  series  r  +  r'-  +  r''  +  •  •  • . 


CONVERGENCE   OF   INFINITE   SERIES  531 


SERIES  WHICH  HAVE  BOTH  POSITIVE  AND  NEGATIVE  TERMS 

General  test  of  convergence.  By  definition,  §  937,  an  infinite 
series  of  any  kind  Wj  +  ^2  +  •  •  is  convergent  if  5„  approaches 
a  finite  limit  as  n  is  indefinitely  increased. 

But,  §§  195,  197,  S'„  will  approach  a  limit  if  the  sequence  of 
values  through  which  it  runs  as  n  increases,  namely,  S-^,  S^, 
S3,  •■•,  possesses  the  property  that  for  every  given  positive 
number  8,  however  small,  a  corresponding  term  S^.  can  be  found 
which  differs  numerically  from  every  subsequent  term  S^.^^  by 
less  than  8.  If  this  condition  is  not  satisfied,  .S^„  will  not 
approach  a  limit,  §  198. 

Since  .s:^.  =  k^  +  . . .  +  w^,^ 

and  ,S\.+^  =  «i  -I f-  Wt  +  i'k+i-\ +  %+p, 

we  have  S^+^  -  5^.  =  ?<i.  +  i  +  w^.+j  +  ■  •  •  +  ^t+p. 

Hence  the  following  general  test  of  convergence  : 

Any  infifiite  series  Ui  +  Ug  +  •  •  •  ^s  convergent  if  for  every 
given  jJositive  iiumber  S,  however  small,  one  can  find  a  term  u^ 
such  that  the  sum  of  any  number  of  the  terms  after  u^  is  numer- 
ically less  than  8;   in  other  words,  such  that 

|%  +  i  +  "k  +  2H l-Uk+pl<  8 

for  all  values  of  p.     If  the  series  does  not  possess  this  property, 
it  is  divergent. 

Hence  in  particular,  a  series  ?<i  +  Wj  +  •  •  •  cannot  be  con- 
vergent unless  lim  «„  =  0.  But  this  single  condition  is  not  suffi- 
cient for  convergence.  We  must  also  have  lim  (ii^  +  w„+i)  —  0, 
lim(«„  +  ?/„_^i  +  ^^,  +  2)  —  0,  and  so  on. 

Thus,  1  +  1/2  +  1/3  +  •  •  •  is  divergent  altliough  lira  u„  =  lim  1/n  =  0. 

For  in  this  series  the  sum  of  the  k  terms  which  follow  the  term  \/k  is 
always  greater  than  1/2.     Thus, 

1  1  1  1      ,     ^      ,  *     7  *  •       ^    1       ;         1 

. 1 1-  ■  •  •  -I >  —  -1{ h  •  •  •  to  A;  terms,  i.  e.  >  —  •  A:  or  - . 

k  +  \      k  +  2  k  +  k     2k      2k  2  k  2 

Hence  k  cannot  be  so  chosen  that  u^+i  +  •  •  •  +  wt  +  t  is  less  than  every 

assignable  number,  and  the  series  is  divergent  (compare  §  948,  2). 


532  A    COLLEGE   ALGEBRA 

956  Corollary  1 .  A  series  ivhich  has  both  positive  and  negative  terms 
is  convergent  if  the  correspo?iding  positive  series  is  convergent. 

For    let   «i  +  ifo  +  ■  ■  ■  (1)    be    the    given    series,    and    let 

u\  +  u'2  H (2)  be  the  same  series  with  the  signs  of  all  its 

negative  terms  changed.     Then 

|%+i  +  «/i-+2  H +  %+p|  <  «'i  +  i  +  «'i-  +  2  -\ h  w'i+p- 

Hence,  if  by  taking  k  great  enough  we  can  make 

«',+  !   +   ■••   +   «',  +  ;.<   8, 

the  same  will  be  true  of  l^  +  i  H h  i<i:+p\-     Therefore  (1) 

is  convergent  if  (2)  is,  §  955. 

957  The  preceding  demonstration  also  shows  that  a  series 
^1  +  ^2  +  •  •  •  with  imaginary  terms  is  convergent  if  the  series 
whose  terms  are  the  absolute  values  of  ?«i,  v^,  •  •  •,  §  2.32,  namely, 
the  series  |«i|  +  |?<2|  +  •  ■  ■>  is  convergent. 

Thus,  t/1  +  iV22  +  ■i-V32  +  ...  is  convergent  since  1  +  1/22  +  1/32  ^.  . . . 
is  convergent. 

958  Corollary  2.  A  series  whose  terms  are  alternatehj  positive  and 
negative  is  convergent  if  each  term  is  numericalhj  less  than  the 
term  which  precedes  it,  and  if  the  limit  of  the  nth  term  is  0. 

For  let  the  series  be  ^i  —  02  +  "s  —  •  •  •,  where  «i,  a.,,  •  •  •  are 
positive.     Using  the  notation  of  §  955,  we  here  have 

I%+i  +  '«i  +  2  H h  ^'i+,.|  =  |«i+i  -  «A-  +  2  H +(-  l)''~'oi.+,.|. 

We  can  write     a^.  +  i  —  a^.^^_  -\ f-  (—  1)^"'%.^^  (1) 

in  the  form  {n, ^^  -  a,^^)  +  {a,^^  -  a^^,)  +  ■  ■  ■  (2) 

and  in  the  form     r/^_^,  —  (rti_^2  ~  '^i  +  a)  ~  ' '  "•  (2) 

Since  a^.^,  >  a^.^g  >  '''a  +  s  >  •••>  each  of  the  expressions  in 
parentheses  in  (2)  and  (3)  is  positive.  Hence  it  follows  from 
(2)  that  (1)  is  positive,  and  from  (3)  that  (1)  is  algebraically 
less  than  %.+  !,  and  therefore  from  (2)  and  (3)  combined  that 
(1)  is  numerically  less  than  %+*• 

But  since  lim  a„  =  0,  we  can  choose  h  so  that  a^.  +  i  <  8. 
Therefore  «!  —  a^  +  a^  —  •  •  •  is  convergent,  §  955. 


CONVERGENCE    OF    INFINITE    SERIES  533 

Absolute  and  conditional  convergence.     A  conyergent  real  series     959 
is  said,  to  be  absolutely  converfjent  if  it  continues  to  be  convergent 
when  the  signs  of  all  its  negative  terms,  if  any,  are  changed  ; 
conditionallij  convergent  if  it  becomes  divergent  when  these 
signs  are  changed. 

Thus,  1  —  1/2  +  1/4  —  1/8  +  •  •  •  is  absolutely  convergent  since  the 
series  1  +  1/2  +  V-4  +  1/8  +  •  •  ■  is  convergent. 

But  1  —  1/2  4-  1/3  -  1/1  +  •  ■  • ,  which  is  convergent  by  §  958,  is  only 
conditionally  convergent  since  1  +  1/2  +  1/3  +  1/4  +  ■  •  •  is  divergent. 

Theorem.     In  an  absolutely  convergent  series  the  positive  terms     960 
hy  themselves  form  a  convergent  series,  and  in  like  manner  the 
negative  terms  by  themselves.    And  if  the  sums  of  these  two  series 
be  P  ayid  —  N  respectively,  the  sum,  of  the  entire  series  is  P  —  N. 

But  in  a  conditionally  convergent  series  both  the  series  of  posi- 
tive terms  and  the  series  of  negative  terms  are  divergent. 

Por  let  Vi  +  ?<2  +  •  •  •  be  a  convergent  series  which  has  an 
infinite  number  of  positive  and  negative  terms. 

Of  the  first  n  terms  of  this  series  suppose  that^j  are  positive 
and  q  negative.  Then  if  S,^  denote  the  sum  of  all  ?t  terms,  Pp 
the  sum  of  the  p  positive  terms,  and  —  N^  the  sum  of  the  q 
negative  terms,  we  shall  have  6'„  =  P^  —  N^. 

When  n  is  indefinitely  increased  both  p  and  q  will  increase 
indefinitely,  and  since  S,^  will  approach  the  finite  limit  S,  one 
of  the  following  cases  must  present  itself,  namely,  either  (1) 
both  Pp  and  N^  will  approach  finite  limits  which  we  may  call  P 
and  N,  or  (2)  both  P^  and  N^  will  approach  infinity. 

In  the  first  case  lim  .§„  =  lim  (P^^  —  N^  =  lim  P^  —  lim  N^, 
§  203,  that  is,  S  =  P  —  N.  The  series  is  absolutely  convergent. 
In  fact,  after  the  change  of  the  signs  of  the  negative  terms  the 
sum  of  the  series  is  P  +  A^. 

In  the  second  case  the  series  is  conditionally  convergent. 
Por  if  S'„  denote  the  sum  of  the  first  n  terms  of  the  series 
obtained  by  changing  the  signs  of  the  negative  terms,  we  have 
lim  5'„  =  lim  {Pp  +  N^)  =  c^. 


534  A   COLLEGE   ALGEBRA 

961  Corollary.     The  terms  of  a  conditionally  conx'erge7it  series  may 

be  so  arranged  that  the  sum  of  the  series  will  take  any  real 
value  that  may  he  assigned. 

For,  as  just  shown,  in  a  conditionally  convergent  series 
the  positive  terms  by  themselves  and  the  negative  terms  by 
themselves  each  constitute  a  divergent  series  the  limit  of 
whose  nth  term  is  0. 

Hence,  for  example,  if  we  assign  some  positive  number  c,  and 
then,  without  changing  the  relative  order  of  the  positive  terms 
or  that  of  the  negative  terms  among  themselves,  form  .§„  by  first 
adding  positive  terms  until  the  sum  is  greater  than  c,  then 
negative  terms  until  the  sum  is  less  than  c,  and  so  on  indefi- 
nitely, the  limit  of  this  S„,  as  n  is  indefinitely  increased,  will  be  c. 

Hence  the  commutative  law  of  addition  does  not  hold  good 
for  a  conditionally  convergent  series. 

EXERCISE   LXXXVII 

1.  Determine  whether  the  following  series  are  convergent  or  divergent. 

n^  3      3  •  5      3  ■  5  •  7       .'>.  •  5  •  7  •  9 
^''3      3-63-6-9      3-6-9.12 

2.  For  what  real  values  of  x  are  the  following  series  convergent  and 
for  what  values  are  they  divergent  ? 

(i)-J-  +  -J-  +  -^L_  +  ...  + '- +  .... 

^^l_x      l  +  2x      l-3x  l  +  (-l)«nx 

(2)-^  +  ^l-  +  -^^  +  ---+     ^"'"'     +■••. 
^  ^  l  +  x2      1  +  2x*      1  +  3x6  l  +  nx'-'" 

3.  If  ii\ -{-  Ui -V  v-z  +  ■  •  •  is  absolutely  convergent,  and  Oi,  a2,  as,  •  •  • 
denote  any  sequence  of  numbers  all  of  which  are  numerically  less  than 
some  finite  number  c,  prove  by  the  method  of  §  956  that  the  series 
aiMi  +  aiU2  +  azUz  +  •  •  •  is  also  convergent. 

4.  If  .S  denotes  the  sum  of  a  series  of  the  kind  described  in  §  958,  show 
that  the  sums  ai,  ai  —  02,  ai  —  a^  +  a^,  ■■  ■  are  alternately  greater  and 
less  than  S. 


convergp:nce  of  infinite  series        535 


CONVERGENCE  OF  POWER   SERIES 

Power  series.     This  name  is  given  to  any  series  which  has     962 
the  form  «o  +  «i^  +  <^2^^  +  •  •  •  +  «„»;"  +  •  •  •  (!)>  where  a;  is  a 
variable  but  a^,  a^,  •  ■■  are  constants.     The  values  of  x  and 
Uq,  Ui,  •  ■  •  may  be  real  or  imaginary. 

By  §  957,  the  series  (1)  is  convergent  if  the  positive  series 
lao|  +  |«ix|  +  [a^x^l  +  •  •  •  +  |«„a;"|  +  •  •  •  (2)  is  convergent.  When 
(2)  is  convergent  we  say  that  (1)  is  absolutely/  convergent 
(compare  §  959).  Whether  (1)  is  convergent  or  divergent  will 
depend  upon  the  value  of  x.  Hence  the  importance  of  the 
following  theorems. 

Theorem  1 .    If  tvhen  x  =  b  evenj  ter'm  of  slq  +  ^iX  +  •  •  •  w     963 

numerically    less    than   some  finite  positive    number  c,   when 
x|<[b|  the  series  is  absolutely  convergent. 

For  since  [  a,p"  |  <  c  for  every  n, 


have  \a,^x"  \  =  |  a,p'"  |  •  y    <  A 


for  every  n. 


Hence  each  term  of  |ao|  +  l^i^;]  +  |a2*^ 


the  corresponding  term  of  c  +  c 


+  •  •  •  (1)  is  less  than 
+  •••(2). 


But  (2),  being  a  geometric  series,  converges  when  \x/h  |  <  1, 
that  is,  when  \x\  <  \b\.  And  when  (2)  converges,  so  does  (1), 
§  945,  1. 

Thus,  l  +  2x  +  x2  +  2x3  +  ---  converges  when  [x]  <  1. 

Corollary  1 .    If  Sio  -\-  aiX  +  ■  •  •  is  converge7it  when  x  =  b,  it  is     964 
absolutely  convergent  when  |x|  <  |b|. 

This  follows  immediately  from  §  963.    For  since  ao  +  «i^  -\ 

is  convergent  when  x  =  b,  all  its  terms  have  finite  values  when 
X  =  b. 

Corollarj^  2.    If  slq  -}-  ajX  +  •  •  •  is  divergent  when  x  =  b,  it  is     965 
also  divergent  when  [x|>|bl. 


536  A    COLLEGE   ALGEBRA 

Por  were  a^  +  a^x  ^ to  converge  for  a  value  of  x  which 

is  numerically  greater  than  h,  it  would  also  converge  for  x  =  b, 
§964. 
966         Limits  of  convergence.     It  follows  from  §  §  964,  965  that  if 
we  assign  to  a  class  A^  all  positive  values   of  x  for  which 

fffl  +  ajx  H converges  and  to  a  class  A^  all  for  which  it 

diverges,  each  number  in  .4i  will  be  less  than  every  number 
in  A2.  Hence,  §  159,  there  is  either  a  greatest  number  in  A^  or 
a  least  in  A^.  Call  this  number  X.  It  represents  the  limit  of 
convergence  of  a^  -\-  a^x  +  ■  ■  ■ ,  the  series  being  absolutely  con- 
vergent when  |ji'l  <  A,  divergent  when  |x-|  >  A. 

Thus,  in  both  x  +  xV2  +  x^/Z  +  •••(!)  and  x  +  x'V^^  +  x^/Z^  +  •  •  •  (2) 
the  limit  of  convergence  \  is  1.  Observe  that  (1)  diverges  and  (2)  con- 
verges when  X  =  X  =  1.  It  is  possible  to  construct  a  series  in  which 
X  =  0  ;  for  example,  the  series  x  +  2  !  x^  +  3  !  x^  +  . .  •. 

What  we  have  called  the  limit  of  convergence  is  more  frequently  called 
the  radius  of  the  circle  of  convergence.  For  if  we  picture  complex  num- 
bers by  points  in  a  plane  in  the  manner  described  in  §  238  and  draw 
a  circle  whose  center  is  at  the  origin  and  whose  radius  is  X,  the  series 
ao  +  aix  +  •  •  •  will  converge  for  all  values  of  x  whose  graphs  lie  within 
the  circle,  and  it  will  diverge  for  all  values  of  x  whose  graphs  lie  without 
the  circle,  §  239. 

967         Theorem  2.     If  in  ao  +  ajX  +  •  ■  ■  the  ratio  \  a„/a„  +  j  [  approaches 
a  definite  limit  fi,  tlien  p,  is  the  limit  of  convergence. 

For,  by  §  952,  the  series  |r/o]  +  |rti./'l  -\ converges  when 

lim     "^''', —  <  1,  that  is,  when  l.rl  <  lim  -^  • 

Similarly  |«„|  +  |aix-|  H diverges  when  \x\  >  lim  M^  • 

Example  L     Find  the  limit  of  convergence  of  the  series 

5         5    10  5- 10- •• 5ft 

bmce  = = — ,  wo  have  fi  =  lim  —  =  -. 

a„  +  i        2n  +  3        2  +  S/n  2  +  3/n      2 


CONVERGEXCE   OF   IXFIXITE    SERIES  537 

Example  2.     Find  the  limits  of  convergence  of  the  series 

■  •  •  +  23x-3  +  S-'^x-s  +  2  x-i  +  1  +  x/3  +  xV32  +  x^fi^  +  • .  • . 

Here  1  +  x/3  +  x-/3-  +  •  •  ■  Is  a  geometric  power  series  In  x  which 
converges  when  |x|<3,  for  a„/a„  +  i  =  3. 

On  the  other  hand,  2x-i  +  22x-2  +  osx-"^  -|-  ...  is  a  geometric  power 
series  in  x-i  or  1/x  which  converges  when  |x-i|<l/2,  and  therefore 
when  |x|>  2. 

Hence  the  given  series  converges  when  2  <|x|<3. 

Example  3.     For  what  real  values  of  x  will  the  series 

x/(l  +  x)  +  2  xV{l  +  x)2  +  3xV(l  +  x)3  +  . .  .  converge  ? 

This  is  a  power  series  in  x/(l  +  x)  which  converges  when  |x/(l  +  x)  |<  1, 
for  11m  a„/rt„  +  i  =  11m  n/{n  +  1)  =  1. 

But  |x/(l  +  x)|<  1  for  all  positive  values  of  x  and  for  negative  values 
which  are  greater  than  —  1  /2.     Hence  the  series  converges  when  x  >  —  1/2. 

The  binomial,  exponential,  and  logarithmic  series.     We  proceed     968 
to  apply  the  preceding  theorem  to  three  especially  important 
power  series. 

1.  The  exponential  series,  §  990,  namely, 

i  +  f  +  f;  +  -  +  n^  •■ 

1       i !  nl 

is  convergent  for  all  finite  values  of  x. 

■r^      ,  a„         1  1 

i  or  here  = = =  n  +  1. 

a„+i      n\      (ri  +  1)! 

Hence  11m  — '—  =  11m  {n  +  1)  =  oo,  that  is,  ^  =  oo. 

O-n  +  l 

2.  The  logarithmic  series,  §  992,  namely, 

J,        6  ^        ^         n 

is  convergent  when  |a'j  <  1,  divergent  when  |xj  >  1. 

^      ,  dn  1  1  n  + 1 

if  or  here      = ; = ■ 

a„  + 1  n      n  +  1  n 

Hence  lim  — ~  =  -  lim =  -  lim  (  \  -\-  -\  =  —  \^  that  is.  u  =  1. 

(1,1  + 1  n  \         11/  '  ' 

The  series  converges  when  x  =  1,  §  958,  diverges  when  x  =  —  1,  §  948. 


538  A    COLLEGE   ALGEBRA 

3.    The  binomial  series,  namely, 

where  m  is  not  a  positive  integer,  is  convergent  when  \x\<  1, 
divergent  when  |a;|  >  1. 

For  here 

a,,    _m  {m  —  \)  •  •  ■  {m  —  n  +  \)      m  {m  —  1)  •  •  •  (m  —  n)  _  n  +  1 

a„  +  i  ~  l-2---n  l-2---(n  +  l)        "  m  -  n 

TT  1-        ^n         ,•       J^  -*■  ^  ,.       1  +  l/n  1    ^,    ^  •  1 

Hence  lim =  lira  — =-i-  --  —  lim =  —  1,  tkat  is,  /u  =  L 

a„  +  1  m  —  11  1  —  m/n 

When  X  =  1  the  series  converges  if  m>  —  1,  diverges  if  ?n <  —  1  (see 

§  1001,  Ex.  2). 

When  X  =  —  1  the  series  converges  if  m  >  0,  diverges  if  m  <  0. 

For  when  x  =  —  1,  by  setting  m  =  —  a  we  may  reduce  the  series  to 

the  form 

a(a  +  l)      a(a  +  l)(a  +  2) 

1-2  1-2-3 

Evidently  from  a  certain  term  on  all  the  terms  are  of  the  same  sign,  so 
that  the  test  of  §  954  is  applicable,  §  956. 

But  here  Un^  ^a  +  n -I  ^n  +  (a  -  \)  _ 

u„  n  n 

Hence,  §  954,  the  series  converges  if  —  (a  —  1)  >  1,  that  is,  if  —  a  >  0, 
or,  since  —  a  =  m,  it  converges  if  m  >  0.  But  it  diverges  if  —  (a  —  1)  <  1, 
that  is,  if  m  <  0. 

EXERCISE   LXXXVIII 

Determine  the  limits  of  convergence  of  the  following  series. 

1.  1  +  mx  +  ??i2xV2!  +  ?n3xV3  !  +  •••. 

2.  2(2x)2  +  3{2x)''  +  2(2x)*  +  3(2x)5  +  -... 

m  (m  —  2)    ,      m  (m  ■—  2)  (m.  —  4)    „ 

3.  mx  +  —^^ X'-  +  — ^ x3  +  . . . . 

For  what  real  values  of  x  will  the  following  series  converge  ? 
3x         1/   3x    \2     1/    3x    \3 


x  +  4   2Vx  +  4/   3\x  +  4/ 

X2  +  1    \X2  +  1/    Vx2  +  1/ 


■(3x)-3+  (3x)-2  +  (3x)-i  +  l  +  2x  +  (2x)2  +  (2x)8  +  .-.. 


OPERATIONS  AVITH    INFINITE    SERIES  539 

XXXIII.    OPERATIONS  WITH  INFINITE  SERIES 

SOME  PRELIMINARY  THEOREMS 

When  a  given  power  series  a^  +  r/^.r  +  •  •  •  is  convergent,  its  969 
sum  is  a  definite  function  of  x  which  we  may  represent  by 
f{x),  writing  f(x)  =  «o  +  a^x  +  •  •  •.  In  what  follows  when 
we  write /(a")  =  a^  -\-  a^x  +  •  •  •,  we  assume  that  a^  +  a^x  +  •  •  • 
has  a  limit  of  convergence  X  whicl  "is  greater  than  0,  and 
suppose  that  \x\<  A. 

Theorem  1 .     Give7i  that  ^  (x)  =  aiX  +  aaX^  + . . . ,  and  that  when     970 
X  has  the  j^ositive  value  b  every  term  of  <^(x)  is  numerically 
less  than  some  finite  positive  number  c. 

If  any  positive  number  8  be  assigned,  hoivever  small,  then 

l</>(x)|  <  8,  ivheJiever  |x|  <  b8/(c  +  8). 

For,  as  was   shown  in  the   proof  in   §  963,  when  [a:;|<  J, 


\4.{x)\<c 
and  therefore  <  c 


+  c 


^  that  is,   <  /^r     §  704 


b\i-\x/b\  '   ^-f1 


Hence  I <^ (a-) I  <  8  when   /'^l  ,  <  8, 

'    ^  ^'  ^  ~  fI 

that  is,  when  kl  < s* 

'     '         C  +  6 

Corollary.     7/"  f  (x)  =  ao  +  aiX  H ,  th  en  lim  f  (x)  =  ao  =  f  (0).     971 

For,  as  just  shown,  li]i^  {a^x  +  a.x^  -\ )  =  0,  §  200. 

Theorem  2.    If  the  series  ag  +  a^x  +  ■  •  ■  vanishes  for  every     972 
value  of  X  for  which  it  cojiverges,  then  0,^  =  0,  ai  =  0,  •  •  •. 

For  setting  x  =  0,  we  at  once  have  a^  =  0, 

Hence  a^x  +  a^x^  +  a^x^  -|-  •  •  •  =  0  ^1) 

for  every  value  of  x  for  which  it  converges. 


540  A   COLLEGE    ALGEBRA 

If  a;  =?^  0,  we  may  divide  (1)  throughout  by  x. 

Hence  a-^  +  Og^  +  "^s^*^  +  •  •  ■  =  0  (2) 

for  every  value  of  x  for  which  it  converges,  except  perhaps 
for  X  —  *d.  But  it  follows  from  this  that  a^  =  0 ;  for  were 
fi^  =^  0  we  could  choose  x  so  small  (without  making  it  0)  that 
|rt2«  +  «3"P^  +  •••!<  [oi|,  §  970,  and  such  a  value  of  x  would  not 
satisfy  (2),  as  we  have  just  shown  it  must. 

Hence  fti  =  0.  And  by  the  same  reasoning  it  may  be  shown 
that  «2  =  0,  «3  =  0,  and  so  on. 

The  like  is  true  of  a  +  hx^  +  ex  +  dx'  H ,  and  of  every  series  in 

which  the  exponents  of  x  are  positive  and  different  from  one  another ;  for 
the  reasoning  just  given  applies  to  all  such  series. 

The  hypothesis  that  Oq  +  aiX  H vanishes  for  every  value  of  x  for 

which  it  converges  contains  more  than  is  required  for  the  proof  that 
Co  =  0,  ai  =  0,  •  •  • .  For  the  reasoning  above  given  shows  that  if  /3i, 
/32j  •  •  •  J  ?m  ■  •  •  denote  any  given  never-ending  sequence  of  numbers  such 
that  lim  /3„  =  0,  and  if  ao  +  aiX  4-  •  •  •  vanishes  when  x  =  )3i,  ^2,  •  •  • ,  /3n,  •  •  • , 
then  ao  =  0,  ai  =  0,  •  ■  • .  In  particular,  the  numbers  Pi,  ^2,  ■  •  ■  may  all 
be  rational. 

973         Theorem  3.    If  ^0  +  ^ix  +  ajx^  -| =  bo  +  bjx  +  bjX^  -| 

for  evert/  value  of  x  for  which  these  series  converge,  the  coefficients 
of  the  likepoivers  ofx  are  equal;  that  is,  slq  =  b^,  aj  =  bj,  a.^  ■=  bj, 
and  so  on. 

For  subtracting  the  second  series  from  both  members  of  the 
given  equation,  we  have,  by  §  974, 

(ao  —  h)  +  («!  —h)x+{a^-b^)x'^-\ =  0 

for  every  value  of  x  for  which  the  given  series  converge. 

Hence,  §  972,  a^-b^^  0,  «i  -  h^  =  0,  a.-b^  =  0,---, 
that  is,  Oq  =  ^o>  «i  =  ^u  ''2  =  ^2>  •  •  •• 

This  theorem  is  called  the  theorem  of  undetermined  coeffi- 
cients. It  asserts  that  a  given  function  of  x  cannot  be  expressed 
in  more  than  one  way  as  a  power  series  in  x  (compare  §  421). 


OPERATIONS   WITH  INFINITE    SERIES  541 

OPERATIONS    WITH   POWER    SERIES 

Since  many  functions  of  x  can  be  defined  by  means  of  power 
series  only,  it  is  important  to  establish  rules  for  reckoning 
with  such  series.  These  depend  upon  the  following  theorems, 
§§  974,  976,  which  we  shall  demonstrate  for  infinite  series  in 
general. 

Theorem    1.     If   the    series   xi^  +  u-z  +  •  ■  ■    end   Vj  +  Vj  +  •  •  •     974 
converge  and   have  the  sums  S  atid   T   respectively,    the  series 
(ui  +  Vi)  +  (u2  +  Vo)  +  •  ■  •   converges  and  has   the  surn  S  +  T. 

For,  §  203,    lim  [{u^  +  v,)  +  (uo  +  v^)  +  •  •  •  +  (»„  +  O] 

=  lim  {ui  +  iin^ h  11,;)  +  lim  (;'i  +  v^ -] 1-  -?;„) 

=  S+T. 

The  like  is  true  of  the  series  obtained  by  adding  the  corre-     975 
sponding  terms  of  any  J7?iite  number  of  infinite  series.    Hence 
the  rule  for  adding  any  finite  number  of  functions  defined  by 
power  series  in  x  is  to  add  the  corresponding  terms  of  these 
series,  that  is,  the  terms  which  involve  like  powers  of  x. 

Thus,  if /(x)  =  1  +  X  +  x2  +  • . .  and  <p  (x)  =  x  +  2  x-  +  Sx^  +  ■  •  ■ , 
then  /(x)  +  0(x)  =  l  +  2x  +  3x2  +  4x3  +  ... 

when  the  given  series  converge,  that  is,  when  |x|<  1. 

If  there  be  given  an  infinite  number  of  series  whose  sums  are 
S,  T,  •••,  and  we  add  the  corresponding  terms  of  these  series, 
we  ordinarily  obtain  a  divergent  series,  even  when  the  series 
.S'  +  7'  +  •  •  •  is  convergent.  But  in  the  case  described  in  the 
following  theorem  we  obtain  a  convergent  series  by  this  process, 
and  its  sum  is  5  +  T"  +  •  •  • . 

Theorem  2.     Let  Ui  +  Uj  +  •  •  •  denote  a  convergent  series  each     976 
of  ivhose  terms  is  the  sum  of  an  absolutely  convergent  series, 
namely, 

Uj  =  u^P  +  u<'>  +  ■  •  •  (1),     U.  =  u^P  +  u<|>  +  •  •  •  (2),  •  • .. 


542  A   COLLEGE   ALGEBRA 

Again,  let  U/,  U2',  •  •  •  denote  the  sums  of  the  series  obtained 
by  replacing  the  terms  of  (1),  (2),  ■■■by  their  absolute  values, 
so  that 

and  so  on. 

If  the  series  Uj'  +  Ug'  +  •  •  •  is  co7ivergent,  the  several  series 
obtained  by  adding  the   corresponding  terms    of  (1),   (2),  •••, 

namely,  the  series  u<}>  +  u^f)  ^ ,  u^p  +  u(|)  -i ,  and  so  07u 

are  convergent,  and  if  their  sums  be  denoted  by  Vi,  Vj,  •  •  •,  we 
shall  have 

Ui  +  U2  +  U3  +  •  •  •  =  Vi  +  V2  +  V3  +  •  •  • . 

For  let  us  represent  the  remainders  after  n  terms  in  the 
series  (1),  (2),  •  •  •  by  R<1\  R^f^,  •  •  -,  §  943,  so  that 

C/i  =  ?<(p  +  ?/p  H +  M^p  +  R^l\ 

U^  =  «<!>  +  w<|>  H h  z<'2)  +  R(i\ 

U^  =  u(\^  +  tf<^  +  •  •  •  +  u^';^  +  R^^\ 


Each  of  the  cohxmn  series  u^\^  +  «*-p  -| ,  «/p  +  w^f)  H , 

and  so  on,  is  convergent  since  each  of  its  terms  is  numerically 
less  than  the  corresponding  term  of  the  convergent  series 
Ui  +  U2+^--,  §945,  1.  Let  the  sums  of  these  series  be 
denoted  by  l\,  l\,  ■■-,  V„,  R„. 

If  we  add  the  corresponding  terms  of  these  n  +  1  column 

series,  we  obtain  the  original  series  Ui  -{-  U^ -\ .     Therefore, 

since  n  is  finite,  we  have,  §  975, 

C^i  +  C^2  +  •  •  •  +  f^t  +  •  •  •  =  Fi  +  F2  +  •  •  •  +  F,.  +  i?,.. 

To  prove  our  theorem,  therefore,  we  have  only  to  show  that 
when  n  is  indefinitely  increased  lim  R„  =  0. 

But  if  the  remainder  after  k  terms  in  7?^^  +  7?^;'  +  •  •  •  be 
denoted  by  6<*>,  we  have  R„  =  R^l^  +  R^-^  -\ h  R^t?  +  ^'^T- 

Let  8  denote  any  positive  number,  it  matters  not  how  small. 
Since  each  term  of  R^]^  +  A";-;'  H (a)  is  numerically  less  than 


OPERATIONS  AVITH    INFINITE    SERIES  543 

the  corresponding  term  of  Ui  +  U^'  +  ■  ■  -(h),  the  remainder 
after  k  terms  in  (a)  is  numerically  less  than  the  corresponding 
remainder  in  (b).  But  since  (b)  is  convergent  we  can  so  choose 
k  that  the  latter  remainder  will  be  less  than  8/2.  Hence  we 
can  so  choose  k  that  ivhatever  the  value  o/n  may  be,  we  shall 
have  S'^l^  <  8/2  numerically. 

But  again,  since  each  of  the  row  series  u^\'  +  u^^^  +  •  •  •  > 
w'^f  +  ?//^' 4- •  •  •,  is  convergent,  as  n  increases  each  of  the  k 
remainders  R^]},  7i^,f ,  •  •  •  R'^^^  will  ultimately  become  and  remain 
numerically  less  than  8/2  k,  and  therefore  the  sum  of  these 
remainders,  namely,  R'^]^  +  iZ^,f  +  •  •  •  +  R'-'i^  will  become  and 
remain  less  than  (8/2  k)k,  or  8/2. 

Therefore,  as  n  increases,  /?„  =  R^l^  +  R^-J  -\ 1-  R^l'>  +  S^^;^ 

will  ultimately  become  and  remain  numerically  less  than 
8/2  +  8/2,  or  8. 

Hence  liui  R,^  =  0,  §  200 ;  and  therefore 

U,+  U^+U,  +  ---=  V,  +  V,  +  F3  +  . . ., 
as  was  to  be  demonstrated. 

A  series  f/i  +  f/2  +  •  •  •  each  of  whose  terms  is  itself  an 
infinite  series  is  called  a  doubly  infinite  series. 

Thus,  consider  the  series 

x/{\  +  x)-  xV(l  +  x)2  +  xy{\  +  x)3 (1) 

which  converges  for  all  real  values  of  x  which  are  greater  than  —  1/2 
(also  for  imaginary  values  of  x  whose  real  parts  are  greater  than  —  1/2). 
Is  it  possible  to  transform  (1)  into  a  power  series  in  x,  that  is,  into  a 
series  which  will  converge  for  any  value  of  x  except  0  ? 

When  |x|<l,  each  term  of  (1)  is  the  sum  of  a  power  series  which  njay 
be  obtained  by  the  binomial  theorem,  §  988.     Thus, 

x/(l  +  x)  =      X  (1  +  x)- 1  =  X  -  x2  +    x3  -    X*  +  •  ■ 

-XV(l+X)2=-x2(l  +  X)-2=        _X2  +  2X3-3X*  +  •• 

xV(l  +  a;)3=      x3(l  +  x)-3=  x3-3x*  +  -- 


(2) 


Replacing  each  term  of  the  first  of  these  series  by  its  absolute  value, 
we  obtain  |x|  +  |x2|  +  \x^\  +  •  ••,  whose  sum  is|xl/(l  —  |xl). 


544  A   COLLEGE    ALGEBRA 

Treating  the  remaining  series  in  a  similar  manner,  we  obtain  series 
whose  sums  are  |x2|/(l  -  |x|)2,  \x^\/{l  -  \x\Y,  and  so  on. 

Hence  the  series  Ui  +  U^'  +  Us'  +  ■  ■  ■  of  our  theorem  is  here 

|X|/(1  -  1X1)  +  |X2|/(1  -  |X1)2  +  |X3|/(1  _  lx|)3  +   •  •  ., 

which  converges  wlien  |x|<  1/2. 

Tlierefore,  when  |x|<l/2,  the  power  series  obtained  by  adding  the 
corresponding  terms  of  the  series  (2),  namely,  x  -  2x2  +  4x3  -  8x*  +  •  •  • , 
converges  and  is  equal  to  the  given  series  (1) ;  that  is,  when  |  x  ]  <  1/2  we  have 

x/(l  +  x)  -  xV(l  +  X)-  +  xV(l  +  x)3 =  X  -  2x2  +  4x^5  -  8x*  +  •  •  •. 

977  Commutative  law  valid  for  absolutely  convergent  series.  We 
are  now  in  a  position  to  demonstrate  that  the  terms  of  an  abso- 
lutely convergent  series  may  he  rearranged  at  jpleasiire  without 
changing  the  sum  of  the  series. 

1.  We  may  rearrange  the  terms  so  as  to  form  any  other 
single  infinite  series  out  of  them. 

For  let  i<i  +  M2  +  •  •  •  (1)  denote  any  absolutely  convergent  series,  and 
let  Ml'  +  M2'  +  •  •  •  (2)  denote  the  same  series  with  its  terms  rearranged. 
Again,  let  8„  denote  the  sum  of  the  first  n  terms  of  (1),  and  S^  the  sum  of 
the  first  m  terms  of  (2). 

Assign  any  value  to  n  ;  then  choose  to  so  that  the  first  n  terms  of  (1)  are 
to  be  found  among  the  first  m  terms  of  (2);  and  finally  choose  p  so  that 
the  first  m  terms  of  (2)  are  to  be  found  among  the  first  n  +  p  terms  of  (1). 

Then  S;„  -  -S„  is  made  up  of  terms  in  S„  +p  -  -S„,  that  is,  of  terms  in 
the  sum  w„  +  i  +  «„  +  2  +  •  •  •  +  u„+p. 

Hence|S;;.-S„|<|u„  +  i|  +  |u„  +  2|  +  ---  +  |u»  +  ;>|. 

But  since  (1)  is  absolutely  convergent,  Hm  (|tt„  +  i]  +  •  •  •  +  |Mn  +  p|)  =  0« 

Therefore  lim  |S;„  -  S„l  =  0,  that  is,  lim  S;„  =  lim  S„. 

2.  We  may  break  the  series  up  into  any  number  (finite  or 
infinite)  of  series  the  terms  of  each  of  which  occur  in  the 
same  order  as  in  the  original  series.  For  we  can  recover  the 
original  series  from  every  such  set  of  series  by  applying  one 
of  the  theorems  of  §§  974,  976. 

Thus,  if  we  form  one  series  out  of  the  terms  of  Mi  +  M2  +  Ws  +  •  •  • 
which  have  odd  indices,  and  another  out  of  those  which  have  even  indices, 
we  have,  by  §  974, 

Wl  +  W2  +  Ma  +  M4  +  •  •  •  =  (Ml  +  M3  +  U5  +  •  •  •)  +  ("2  +  W4  +  "6  +  •  •  •)• 


OPERATIONS  WITH   INFINITE    SERIES  545 

Or  again,  arrange  the  terms  of  Mi  +  U2  +  ?«3  +  •  •  •  as  follows : 
Ml  In  this  scheme  there  are  an  infinite  number  of 
u<2  +  7(3                        columns,  each  forming  an  infinite  series. 
M4  +  Ms  +  We                    The  sum  of  Ui  +  W2  +  M3  +  •  ■  •  is  equal  to  the  sum  of 
W7  4-  "8  +  Mg  +  Uio    the  terms  of  the  scheme  added  by  rows,  §  940.     And 
the  sum  by  rows  is  equal  to  the  sum  by  columns,  §  976. 

Hence  Mi  +  M2  +  M3  H =  ("i  +  W2  +  "4  H )  +  (ws  +  U5  H ) -\ . 

And  similarly  in  every  case. 

3.    Every  possible  rearrangement  of  the  terms  of 

iti  +  «2  +  M3  ^ \-^in-i 

may  be  had  by  combining  1  and  2. 

Products  of  power  series.     If  the  functions/(a3)  and  <f>  (x)  are 

defined,  when  |  •?- 1  <  A,  by  the  power  series/(a;)  =  ao  +  ^1^  H (1)> 

(f,  (a-)  =  ^-0  +  ^'i^'  +  •  •  •  (2),  their  product  f(x)  ■  <^  (x)  will  be 
defined,  when  |a'|  <  A,  by  a  power  series  derived  from  (1)  and 
(2)  by  the  ordinary  rules  of  multiplication  (compare  §  314). 

Thus,  /{x)=ao     +  aix       +  Uox"^       +  aax"       -\ (1) 

0  (X)  =  60     +  hx       +  6ox2       +  63x3       +  •  •  •  (2) 


/(x)  •  <^  (X)  =  ao6o  +  aibo  I  x  +  «2^o  x^  +  a-zK  x^  +  ■  •  •  (3) 


ao6o  +  aiho  x  +  a^ho 

x2  +  azh» 

+  ao6i      +  ai6i 

+  a^hi 

+  aob-i 

+  ai62 

+  ao63 

For  when  |x]<X  so  that  (1)  and  (2)  are  convergent,  we  have,  §939, 
/(x)  0  (X)  =  f{x)  bo  +f{x)  bix  +  fix)  60x2  +  . . . . 

This  is  a  series  of  the  kind  described  in  §  976,  for  it  remains  con- 
vergent when  all  the  terms  of  /(x)  and  <f>{x)  are  replaced  by  their 
absolute  values.     We  may  therefore   add   the   corresponding  terms  of 

/(x)  60  =  aobo  +  aib(yc  H ,  /(x)  6iX  =  0  +  ao&iX  +  •  •  • ,  and  so  on.     The 

result  is  the  series  (3). 

Example.     Express  (1  +  x  +  x2  +  •  •  •)  (1  +  2  x  +  3  x^  +  •  •  •)  as  a  power 


Transformations  of  power   series.     Suppose   that  the  power 

series  rr,,  +  aii/  +  a„y/'^  +  ■■•(!)   converges  when  jy]  <  X. 

Suppose  also  that  y  may  be  expressed  in  terms  of  x  by  the 
power  series  ]/  =  b^ -\- b^x  -\ (2),  where  [6o|  <  A. 


546  A    COLLEGE   ALGEBRA 

By  repeatedly  multiplying  (2)  by  itself,  we  obtain  expres- 
sions for  y,  if',  i/,  •  •  •  in  the  form  of  power  series  in  x  which 
converge  when  (2)  converges.  If  we  substitute  these  expres- 
sions in  the  terms  a^y,  a^^i/'^,  ■  •  •  of  (1),  we  obtain  a  series  of 
the  form  a^  +  a^  (b^  +  h^x  +  ---)  +  a^  {!>%  +  2  hjj.x  +  ...)  +  ...  (3), 
and  this,  when  the  terms  which  involve  like  powers  of  x 
are  collected,  becomes  a  power  series  in  x  of  the  form 
K  +  «A  +  •••)  +  («i^'i  +  2  aAh  +  •  •  Oa-  +  •  •  •  (4). 

This  final  series  will  converge  and  have  the  same  sum 
as  (1)  for  all  values  of  x  such  that  |&„|  +  |Jia-|  +  •  •  ■  <  X. 
For  in  this  case  the  condition  of  §  976  is  satisfied  by  the 
doubly  infinite  series  (3),  the  series  f/j'  +  Uo,'  +  •  ■  •  being 
|ao|  +  |«i|  (l^ol  +  \^i^\  +  •••)+■■■'  which  by  hypothesis  con- 
verges when  l^ol  +  I ^1-^1  +  •  •  •  <  A. 

980         Quotients  of  power  series.     A  fraction  whose  numerator  and 
denominator  are  power  series,  as 

(«o  +  a,x  +  -.  ■)/{]>,  +  b,x  +  ■■■), 

where  ^o  "^  0,  may  be  transformed  into  a  power  series  which 
will  converge  for  all  values  of  x  for  which  a^  +  ciiX  +  •  •  •  con- 
verges and  \hix\  +  \h^x'^\  -f  •  •  •  <  \b^^\. 

For  let  y  =  biz  +  hoz"-  +  •  •  • .  (1) 

1  111 


Then 


bo  +  6ix  +  62x2  +  . . .      60  +  2/      60    1  +  y/bo 


(2) 


60 

since,  by  hypothesis  and  §  232,  |?/|<|6ix|  +  \biX-\  +  ■  •  •  <]6o|- 

In  (2)  replace  y  by  its  value  (1)  and  then  apply  §  979.  We  shall  thus 
transform  (2)  into  a  power  series  in  x  which  converges  when 

|5lX|  +  |?^2X2|+-.-<|6o|. 

Multiply  this  power  series  by  a^  +  ctiX  +  02x2  +  ■••,§  978. 

The  result  will  be  a  power  series  in  x  which  will  converge  and  be  equal 
to  the  given  fraction  for  all  values  of  x  for  which  ao  +  aiX  +  •  •  •  converges 
and|6ix|  +  |62-c21+  •v<|&o|- 


1  +  3  +  7  +  ■  •  • 
1  +  1  +  1  +  --- 

2  +  6  +  --- 
2  +  2  +  • .  • 

OPERATIONS  WITH   INFINITE   SERIES  547 

The  quotient  series  may  be  obtained  to  any  required  term 
by  the  process  of  cancelling  leading  terms  described  in  §  406, 
or  by  the  method  of  undetermined  coefficients,  §  408. 

Example.  Expand  (1  +  2  x  +  2^2  +  . .  .)/(i  ^  x  +  x^  +  ■  •  ■)  to  four 
terms. 

Using  detached  coefficients,  we  have 

1  +  2  +  4  + 8  + ■■•  11  +  1  +  1  +  1  +  ... 
1  +  1  +  1  +  1  +  ■  •  •  |l  +  l  +  2  +  4  +  ... 

Hence  the  quotient  series  is 

1  +  a;  +  2  x2  +  4  x3  +  .  . . . 

It  converges  when  |x|  <  1/2. 

4  +  ... 

When  instead  of  being  infinite  series  the  numerator  and  981 
denominator  are  polynomials  in  x,  so  that  the  fraction  has  the 
form  (^?,)  -\-  a^x  -{-■■•  +  ci',n^'")/(^o  +  biX  +  ■  •  ■  -\-  b,flf),  the  quo- 
tient series  will  converge  for  all  values  of  x  which  are  numer- 
ically less  than  the  numerically  smallest  root  of  the  equation 
h^  +  h^x  +  •  ■  ■  -\-  h^x"  =  0.  This  will  be  evident  from  the  first 
of  the  following  examples.  The  second  of  these  examples 
illustrates  the  form  of  the  quotient  series  when  b^  =  0,  and  the 
third  illustrates  the  method  of  expressing  the  fraction  as  a 
power  series  in  1/x. 

Example  1.  Find  the  limit  of  convergence  of  the  series  which  is  the 
expansion  of  (3x  +  8)/(x2  +  .5x  +  6). 

By  the  method  of  partial  fractions,  §  537,  we  find 

3x  +  8  3x  +  8  21 


and 


x2  +  5  X  +  6      (x  +  2)  (X  +  3)      X  +  2      x  + 

_2_ 
xT2      1  +  x/2 

1 


But-^  =  — ^- —  =  ^1  +  -")    '=1--  +  -" when|xl<2, 

-x/2      \         2/  2      22  II' 


X  +  3      3  1  +  x/3 


=  Vl+^^    '^l_^  +  ?! when|xl<; 

3  \        3  /  3      32      33  '    ' 


^^      r  3x  +  8  4      llx      31x2  V      ,    ,^0 

Therefore = when  x  <  2. 

X2  +  5X  +  6      3        18         108  '    ' 


548  A   COLLEGE    ALGEBRA 

Example  2.     Expand  (1  -  x)/(x-  +  4  x^)  in  increasing  powers  of  x. 

We  have  ]  ~f  ^  =  l-l::!^  =  1(1  -  5x  +  20x2  -  80x3  +  •  ■  •) 

x2  +  4x3      xn  +  4x      x^^  ' 

=  x-2  -  5x-i  +  20  -  80x  +  •  •  ■. 
Example  3.     Expand  (2x2  +  x  -  3)/(x3  +  2x  +  4)  in  powers  of  1/x. 
2  x2  +  X  -  3  _  1    2  +  1/x  -  3/x2  _  1  /q      3       5        \_2_3        5 
x3  +  2x  +  4~x'l  +  2/x  +  4/x3  ~x\        x      ^")~x      x^~x^"'' 

982  Reversion  of  series.  From  the  equation  j/  =  a^x  +  a^x^  +  •  •  •> 
defining  y  in  terms  of  x,  it  is  possible  to  derive  another  of  the 
form  X  =  b^y  +  b^ij^  +  •  •  •>  defining  x  in  terms  of  y.  The  pro- 
cess is  called  the  reversion  of  the  given  series  a^x  +  a2X^  +  •  •  •• 
It  will  be  observed  that  this  series  lacks  the  constant  term  a^, 
and  the  understanding  is  that  a-^  ^  0.  It  can  be  proved  that 
if  aiX  -f  Uox'^  +  •■•  has  a  limit  of  convergence  greater  than  0, 
the  like  is  true  of  the  reverted  series  h^y  +  b^y'^  -\ . 

Example.     Revert  the  series  y  =  x  +  2x'^  +  Zx^  -\-  ■  ■  • . 
Assume  x  =  feiy  +  h-zy-  +  h-nr  +  ■■■.  (1) 

Computing  y'^,  y^,  ■  ■  ■  from  the  given  equation  by  the  method  of  §  978, 
and  substituting  the  resulting  series  in  (1),  we  have 

x3  +  . .  .  (2) 


M  + 

2  6i 

x2  +  3  6i 

+ 

h 

+  46, 
+     h 

Equating  coefficients,  1  =  6i,  0  =  2  6i  +  60,  0  =  3  61  +  4  62  +63,  •  •  • 
whence  61  =  1,  60  =  —  2,  63  =  5,  •  •  • . 

Therefore  x  =  y  -  27/^  +  [>y^  +  ■  ■  ■.  (3) 

By  the  same  method,  from  an  equation  of  the  form 
y  =  aQ-\-  a^x  -f  a^x'  +  •  •  ■ ,  or  //  —  a^J  =  (tiX  -\-  a„x'  +  •  •  ■ ,  we  can 
derive  another  of  the  form  x  =  b^  (y  —  r?„)  +  b„  (y  —  a^'^  +  •  •  • . 
And  from  an  equation  of  the  form  y  =  a^x'  +  a.^x^  +  •  ■  •  we 
can  derive  two  others  of  the  form  x  —  b^y^  +  b^y  +  b.^y^-  +  •  •  •. 

Expansion  of  algebraic  functions.  An  algebraic  equation  of 
the  form  f{x,  y)  =  0  which  lacks  a  constant  term  is  satis- 
fied when  cc  =  0  and  y  =  0.     Hence,  if  we  suppose  f{x.,  y)=  0 


OPERATIONS  WITH   INFINITE    SERIES  549 

solved  for  y  in  terms  of  ar,  one  or  more  of  the  solutions  must 
be  expressions  in  x  which  vanish  when  x  vanishes.  It  can 
be  proved  that  these  expressions  may  be  expanded  in  series 
in  increasing  powers  of  x  which  have  limits  of  convergence 
greater  than  0.  In  ordinary  cases  these  series  may  be  obtained 
to  any  required  term  by  the  method  illustrated  in  the  follow- 
ing examples. 

Example  1.     The  equation  7/2  +  ?/  — 2x  =  0  lacks  a  constant  term. 
Find  the  expansion  for  the  value  of  y  which  vanishes  when  x  =  0. 
When  X  =  0,  the  equation 

2/2  +  y  -  2  X  =  0  (1) 

becomes  7/2  -f  ?/  =  0.  Since  one  and  but  one  of  the  roots  of  this  equation 
is  0,  one  and  but  one  of  the  solutions  of  (1)  for  y  in  terms  of  x  vanishes 
when  X  vanishes. 

Suppose  that  when  this  solution  is  expanded  in  a  series  of  increasing 
powers  of  x  its  first  term  is  ax'^,  so  that 

2/  =  ax'^  +  -.-.  (2) 

Substituting  (2)  in  (1),  we  have 

a2a;2M  + [-  cfx'^  + 2  X  =  0.  (3) 

Since  by  hypothesis  (3)  is  an  identity,  the  sums  of  the  coefficients  of 
its  terms  of  like  degree  must  be  0.     Hence  there  must  be  at  least  two 
terms  of  lowest  degree ;  and  since  \j.  is  positive,  these  must  be  the  terms 
ax'^  and  —  2  x.     Therefore  /^  =  1  and  a  —  2  =  0,  or  o  =  2. 
We  therefore  assume  that 

2/  =  2  X  +  6x2  +  cx^  +  •  •  • .  ^2') 

Substituting  (2')  in  (1),  we  obtain  (4  +  6)x2  +  (4  &  +  c)x3  +  •  • .  =  0. 
Hence  4  +  6  =  0,  46  +  c  =  0,  ■••,  and  therefore  6  =  —  4,  c  =  16,  •  •  • . 
Therefore  the  required  solution  is  y  =  2  x  —  4  x2  +  16  x^  +  •  •  • . 
Example  2.     Find  the  expansions  of  the  values  of  y  in  terms  of  x  which 
satisfy  the  equation  2/3  —  x?/  +  x2  =  0  and  vanish  when  x  =  0. 
When  X  =  0,  the  equation 

2/3  _  a;2/  +  x2  =  0  (1) 

becomes  y"^  =  0,  all  three  of  whose  roots  are  0.  Hence  we  may  expect  to 
find  three  expansions  of  the  kind  required. 

Let  ax^  denote  the  leading  term  in  one  of  these  expansions,  so  that 
2/  =  ax/^  +  ■  •  • .  (2) 

Substituting  (2)  in  (1),  we  have 

aH^>^  +  •  •  ■  -  ax*^  + 1  + f-  a;^  =  0.  (3) 


550  A   COLLEGE    ALGEBRA 

By  the  reasoning  of  Ex.  1,  at  least  two  of  the  exponents  3yu,  yit  +  1, 
and  2  must  be  equal  and  less  than  any  other  exponent  of  x  in  (3). 

Setting  3ja  =  ^  +  1,  we  find  ij.  =  1/2.  This  is  an  admissible  value  of  fj., 
since  when  /j.  =  1/2,  both  3/i  and  fi  +  1  are  less  than  2. 

Setting  ;u  +  1  =  2,  we  find  /x  =  1.  This  also  is  an  admissible  value  of 
/n,  since  when  m  =  1,  both  /x  +  1  and  2  are  less  than  3^. 

Setting  3^  =  2,  we  find  /j.  —  2/3.  But  this  is  not  an  admissible  value 
of  M,  since  when  /x  =  2/3,  3  /a  and  2  are  greater  than  /u  +  1. 

Hence  jit  must  have  one  of  the  values  1  or  1/2. 

When  ,Lt  =  1,  (3)  becomes  o?x^  +  •  •  •  —  ax-  +  •  •  •  +  a;^  =  0,  from  which 
it  follows  that  —  a  +  1  =  0,  or  a  =  1. 

When    At  =  V-i    (3)    becomes    a^x'  +  • . .  -  ax?  H +  x'-^  =  0,    from 

which  it  follows  that  a^  —  a  =  0,  or,  since  a  ?i  0,  that  a  =  ±  1. 

We  therefore  assume  that  the  required  solutions  are  of  the  form 
y  =  X  +  6x'-  +  cx^  H ^  y  =  x^  +  6x  +  cx^  +  .  •  • ,  y  =  —  x^  +  6x  +  ex?  H . 

And  substituting  these  expressions  for  y  in  (1)  and  determining  the 
coefiicients  as  in  Ex.  1,  we  obtain 

,     o  ,  o    ■?  ,  1      X      3x?  1      X      3x? 

7/  =  X  +  X-  +  3x3  +  •..,  2/  =  x^  ------+...,  2/  =  -  x=  --  +  —-  +  ••• . 

2  o  2  o 

In  this  method  it  is  assumed  that  if  the  leading  term  of  one 

of  the  required  expansions  is  ax'' ,  the  expansion  will  be  in 

1 

powers  of  x"^.  In  exceptional  cases  this  is  not  true  and  the 
method  fails.  But  the  following  method  is  general.  Having 
found  the  leading  term  ax'*  of  an  expansion  as  in  the  examples, 
set  y  =  x**  (a  +  v)  in  the  given  equation.  It  becomes  an  equa- 
tion in  V  and  x.  From  this  equation  find  the  leading  term  of 
the  expansion  of  v  in  powers  of  x,  and  so  on.* 

Thus,  in  Ex.  2,  setting  ?/  =  x^  (1  +  ?j)  in 

2/3  -  xy  +  x2  =  0  (1) 

and  simplifying,  we  have 

1)3  +  3  u2  +  2  V  +  x^  =  0  ;  (2) 

whence  r=  —  X-/2H ,  and  therefore  2/=x^(l— xV2H )  =  x'— x/2H , 

To  find  the  next  term,  set  u  =  x^(—  1/2  +  v')  in  (2),  and  so  on. 

*For  a  fuller  discussion  of  the  iiiothods  of  thi.s  section  and  the  use  in  con- 
nection with  them  of  Newton's  parallcloijvain  see  Chrystal's  Alr/ebra,  II, 
pp.  349-371 ;  also  Frost's  Curve  Tracing  and  Johnson's  Curve  Tracing 


OPERATIONS   WITH    INFINITE    SERIES  551 

Taylor's   theorem.      If    f{x)  =  a^  +  a^x  +  a^x^  +  •••    when     984 

ja;|<A,  and  we  replace  x  by  x  +  h,  we  obtain 

fix  +  h)  =  ao  +  Oi(ic  +  h)  Jra^{x  +  hy  +  --  ■. 

It  follows  from  §  976  that  when  |ic|  +  |/il  <  A  we  may  transform 
this  series  into  a  power  series  in  h  by  expanding  (a  +  Jif, 
(a  +  hy,  ■  •  •  by  the  binomial  theorem,  and  then  collecting  terms 
which  involve  like  powers  of  h.  By  the  method  employed  in 
§  848  it  may  be  shown  that  the  result  will  be 

f(x  +  h)=f{x)+f{x)  h  +/"(x)  ■  f-j  +  •  • .  +/")  C-^)  •  5  +  •  •  •, 

where  /'  (a;),  /"  {x),  ■  •  •  denote  the  sums  of  the  series  whose 
terms  are  the  first,  second,  •  •  •  derivatives  of  the  terms  of  the 
given  series  a^  +  «i-^'  +  "a^"^  +  •  •  •>  namely, 

/'  (x)  =  ai-\-2aoX  +  S  a^x^  H , 

/"  (x)  =  2  (12  +  3  -2  a^x  +  •••,  and  so  on. 

If  in  the  preceding  identity  we  replace  x  by  a  and  h  by     985 
X  —  a,  where  |a|  +  |a5  —  a]  <  X,  we  obtain  the  expansion  oi  f{x) 
in  powers  of  x  —  a,  namely, 

f(x)=f(a)+f'(a)(x-a)+...+r(a)^^-^  +  .... 

From    this    last   expansion   and    §  971    it   follows   that   if     986 
/(x)  —  a^-\-  a^x  +  •  •  •  when  |a-|  <  A,  and  if  [a]  <  X,  then 

\\mf(^x)=f(a). 


EXERCISE  LXXXIX 

1.  Show  that  (1  +  X  +  x2  +  . .  ■)2  =  1  +  2  a;  +  3  x2  +  4  z'  +  . .  • . 

2.  Show  that  (1  +  a;  +  x^  +  •  •  •)3  =  1  +  3 x  +  6 x^  +  10 x^  +  . . .. 

3.  Show  that  (1  +  x2  +  x^  +  •  •  ■)/{!  +  z  +  x-  +  ---)  =  l-x-\-  x^-    ■  -. 

4.  Assuming  that  (1  —  x  +  2  x^)^  =  1  +  aiX  +  02x2  +  •  •  • ,  find  ai,  Os, 
as,  Ui  by  squaring  the  given  equation  and  applying  §  973, 


552  A    COLLEGE   ALGEBRA 

5.  By  a  similar  metliod  find  the  first  four  terras  of  tlie  expansions  of 
(l)(8-3x)i  (2)  (l  +  x-x2)l 

6.  Expand  eacli  of  the  following  fractions  in  ascending  powers  of  x 
to  the  fourth  term  by  the  method  of  the  example  in  §  980. 

2  +  x-3x2+  5a;3  x  +  5  x^  -  x^ 

^'       l  +  2x  +  3x2      ■  ^  ^  l-x  +  x-i-xs" 

7.  Expand  each  of  the  following  fractions  in  ascending  powers  of  x 
to  the  fourth  term  by  the  method  of  undetermined  coefl&cients. 

3x2 +  x3  »  +  5x* 

^  '  l  +  a;  +  x2  ^      x3  +  2x*  +  3x5 

8.  Expand  each  of  the  following  fractions  to  the  fifth  term  by  the 
method  of  the  first  example  in  §  981  and  indicate  the  limits  of  con- 
vergence of  the  expansions. 

9X-22  5X  +  6 

^  ^  (x2-4)(x-3)  ^  ^   (2x  +  3)(x  +  l)2 

9.  Expand  each  of  the  following  fractions  to  the  fourth  term  in 
descending  powers  of  x.  For  what  values  of  x  will  the  first  of  these 
expansions  converge  ? 


2  x2  +  X  -  15  x*  +  x3  +  x2  +  X  +  1 

10.  Revert  each  of  the  following  series  to  the  fourth  term. 

(1)  ?/  =  x  +  x2  +  x3  +  x*  +  ..-.      (2)  2/  =  x-|-  +  |--|-  +  --.. 

11.  From  2/  =  1  +  X  +  xV2  +  xV3  +  •  •  •  derive  to  the  fourth  terra  a 
series  for  x  in  powers  of  y  —  1. 

12.  From  y  =  x^  +  3  x^  derive  to  the  fourth  term  a  series  for  x  in 
powers  of  y^. 

13.  By  the  method  of  §  983  find  the  first  three  terms  of  the  expansions 
of  the  values  of  y  in  terms  of  x  which  satisfy  the  following  equations  and 
vanish  when  x  =  0. 

(1)  x2  + 2/2  +  2,  _3x  =  0.  (2)  x8  +  2/3  -  xy  =  0. 

14.  By  aid  of  the  theorem  of  §  976  show  that 

X  2x2  3x3  X  .  X2  .  X8  , 


1  -  X         1  -  X2        1  -  X3  (1  -  X)2         (1  -  X2)2        (1  -  X^f 


THE    BINOMIAL    SERIES 


553 


XXXiy.     THE    BINOMIAL,    EXPONENTIAL, 
AND    LOGARITHMIC    SERIES 


(ffi 


The  binomial  series. 
7)1  (m 


When  VI  is  a  positive  integer, 


98? 


1  +  T--  a-  + 


1-2 


n,.  +  !!Ll!!L 


!)(»'- 2) 


l'J'3 


•'  + 


(1) 


is  a  finite  or  terminating  series  and  its  sum  is  (Iji,^)'". 

When  m  is  not  a  positive  integer,  (1)  is  an  infinite  series, ' 
but  one  vrhich  converges,  that  is,  has  a  sum,  when  \x\<  1, 
§  968.  We  proceed  to  demonstrate  that  if  m  has  any  rational 
value  whatsoever,  this  sum  is  (1  +  or)"'. 

The  series  (1)  is  a  function  of  both  x  and  in,  but  since  we 
are  now  concerned  mainly  with  its  relation  to  m  we  shall 
represent  it  by   </>(»«). 

For  convenience  let  w,.  denote  the  coefficient  of  x^  in  (1),  so 
that  v}y  =  m (rn  —  !)■  ■  •  (in  —  r  +  1)/'" !• 

Then  if  m  and  n  denote  any  two  numbers,  we  have 
(^  (m)  =  1  +  ni^x  +  vux"  +  m^x^  +  ■•■■> 


^  (/;)  z=  1  +  n^x  +  tux'  +  Ji^x^  H , 

<^  (m  +  n)  =  1  +  (m  +  n)iX  +  (??i  +  n).yx'^  H . 

We  can  prove  that  (ft  (w)  •  ^(n)  =  (^(w  +  71). 

For  when  \x\<  1,  so  that  (2)  and  (3)  converge,  we  have 


(-0 

(3) 
(4) 


(5) 


But  in  §§  773,  774  it  is  shown  that 
77?j  4-  nj  =  (m  +  ?/)i,   m„  +  7n^ni  +  w.,  =  (^"  +  n)n,  ■  ■  ■, 
VI,.  +  w^_i"i  +  •  •  •  +  Wi7j,._i  +  71,.  =  (m  4-  w),.. 
Hence  <^  (//;)•  <^  (w) 

=  1  4-  (wi  +  ")i-^  +  i^"-  +  ")2-'''''  +  0^  +  ^Os'^""  H 

=  <t>(m  +  n).  (6) 


<^  (w)  •  <^  (n)  =  1  +  ?«-i  X  ■{-  Tn^ 

X-  +  7^3 

+  Wi       +  ??ii?ii 

4-  wij^^-i 

+  n. 

4-  miWo 

+  W3 

554  A   COLLEGE    ALGEBRA 

By  repeated  applications  of  (G),  we  have 

<l>(m)-cf)(n)  ■<ji(p)=  <f>(m  +  ??)-<^(;/)=  <^(y«,  +  ?i  4- ;y), 

and  so  on,  for  any  finite  number  of  factors  of  the  form  </>  (??«), 
<f>(n),<f>{p),<i>(q),--: 

We  are  now  prepared  to  prove  the  binomial  theorem  for  all 
rational  values  of  the  exponent  ?«,  namely : 

988         Theorem.     //*  m  he  any  rational  number  whatsoever,  the  sum 
of  the  series 

.    s       .       1"          m(m  —  1)    .,       m(m  —  l)(m  —  2)    „  , 
</>(m)=l  +  jX+      \    ,,     'yi'+      ^      ^^;^ '-x'+--; 

when  [x|  <  1,  is  (1  +  x)'". 

Notice  first  of  all  that  when  m  =  0  the  series  reduces  to  1, 
and  that  when  ni  =  1  the  series  reduces  to  1  +  cr. 

Hence  <^(0)=  1  and  <^(1)=  1  +  ir.  (1) 

1.  Let  VI  be  a  positive  integer. 

Then  <^  (ot)  =  <^  (1  +  1  +  •  •  •  to  m  terms) 

=  ^(1)  •  <^(1)-  ••  to  m  factors 
=  [«^(l)]"'  =  (l+:r)"',  by(l),       (2) 

which  proves  the  theorem  for  a  positive  integral  exponent. 

2.  Let  m  be  any  positive  rational  fraction  7) Z^-. 

Then  [<^ {p/q)y  =  4>  {p/q)  ■  ^ (p/'i)  •  •  •  to  ?  factors 

=  i>  ivl'i  +  pIi  H to  ?  terms) 

=  <^(/0  =  (l+x)^  by  (2). 

Therefore  <^  {plq)  =  (1  +  a-)".  (3) 

For  it  follows  from  the  equation  [<^  (/?/!?)]'  =  (^  +  ^Y  ^"^ 
§  986  that  the  values  which  ^{^pl'i)  takes  for  all  values  of  x 
such  that  la*!  <  1  must  be  the  corresponding  values  of  one  and 
the  same  qXh.  root  of  (1  +  a-)  p. 


THE    BINOMIAL   SERIES  555 

Moreover  this  root  must  be  the  principal  qth.  root,  namely, 
p 
(1  +  xy ;  for  this  is  the  only  one  of  the  ^■th  roots  of  (1  +  xy 
which  has  the  same  value  as  ^  {p/q)  when  x  =  0. 

3.    Let  m  be  any  negative  rational  number  —  s. 

Since       <^(- s)  •  <^(s)  =  <^(- 5  +  s)=  <^(0)  =  1,  by  (1) 

we  have  <^  (-  «)  =  l/<^  (s)  =^  1/(1  +  x)'  by  (3) 

=  {l+x)'%  (4) 

which  proves  the  theorem  for  any  rational  exponent. 

It  is  not  difficult  to  extend  the  theorem  to  irrational  values 
of  the  exponent. 

Example.     Expand  (1  +  2  x  +  3  a;^)'  in  ascending  powers  of  x. 
We  have  (1  +  2  x  +  3  x^)^  =  [1  +  (2  x  +  3  x2)]3 

=  1  +  1  •  (2 X  +  3x2)  +  ilzill  (2 X  +  3x2)2 

\(_   2W_5\ 

+  '^    ^[l    ^\2x  + 3x2)3  +  ... 

_         2  X      5  x2      68  x3 
''"T'''~9  81         ' 

The  expansion  converges  when  2|x|  +  31x2|<l; 

therefore  when  9|x2|  +  6  |x|  +  1  <4  ; 
therefore  when  3|x|  +  l<2; 
therefore  when  |x|<l/3. 


Corollary.    If  va.  is  rational  and  |x|  <  |a],  we  have  989 

>  - 
1-2 


1         m(m  —  1)         „    . 
(a  +  x)™  =  a™  +  ma™-^x  +  — \  ^a'"-''x''4- 


For 


r  x       m{m  —  l)x'^  ~\ 


(1) 


=  a'"  +  ma"-ia;  +  '^^'^'  ^  ^^  a^-^x^  +  •  •  •,   (2) 
where  (1)  and  therefore  (2)  converge  if  \x/a\  <  1,  or  [a:[  <  [a|. 


990 


556  A  COLLEGE  ALGEBRA 

The  exponential  series.     We  have  already  shown    that   the 
series 

1  +  x/l+xy2l  +  0:73!  +  •  ••  +  x"/7il  +  •••  (1) 

converges  for  all  finite  values  of  x,  §  968. 

Let  e  denote  its  sum  when  a-  =  1,  so  that 

e  =  l  +  l  +  l/2!  +  l/3!  +  ---  =  2.71828---. 

We  are  to  prove  that  the  sum  of  (1)  for  any  real  value  of  x  is  e'. 

For  letf{x)  denote  the  sum  of  (1),  so  that 

f(x)  =  1  +  :r/l  +  x'/2 !  +  xys  !  +  ••■+  x^/n  !  +  •.., 

/(y)=  1  +  y/1  +  f/2l  +  ,//3!  +  ...  _^  yV«!  +••.. 

Then  by  the  rule  for  multiplying  infinite  series,  §  978, 

/(^)  -Ai/)  =  1  +  (^  +  y)  +  (iJ  +  ^-y  +  f-J) 

\3l^  211^  121^  3lJ^"' 

From  this  result  it  follows  that 
/(^)  -/(y)  •/(«)  =/(^  +  y)  ■f{^)=f{x  +  y-\-z),  and  so  on. 

Hence,  observing  that/(0)=  1  and  /(I)  =  e,  we  may  prove 
successively,  precisely  as  in  §  988,  that 

1.  When  a;  is  a  positive  integer  m, 

2.  When  cc  is  a  positive  fraction  ji/q, 


3.    When  a;  is  a  negative  rational  —  s, 

/(-.)=l//(.)=l/e«  =  e-. 


THE   LOGARITHMIC    SERIES  557 

Therefore  when  x  is  rational  we  have  f{x)  =  e%  that  is, 

e-  =  1  +  X  +  a;V2!  +  ^yS!  +  •  •  •  +  a;"/^!  +  •  •  ••         (2) 

Moreover  (2)  is  also  true  for  irrational  values  of  the  expo- 
nent X.  For  if  h  denote  any  given  irrational  number  and  x  be 
made  to  approach  h  as  limi-t  through  a  sequence  of  rational 
values,  for  all  these  rational  values  of  x  we  have  fix)  —  e"  and 
therefore  ^'^^'^fix)  =  l^n  e^. 

x=b^  ^    ^  x=b 

But  ^i"V(^)  =/(^)'  §  986,  and  lif^  e^  =  e\  §  728.  There- 
fore/(6)~i  e^  that  is,  1  +  ^  +  by 2  !'+•••  =  e\ 

The  second  member  of  (2)  is  also  a  convergent  series,  that  is,  has  a 
sum,  when  x  is  imaginary.  Hence  (2)  may  be  used  to  define  e-'  for  imagi- 
nary values  of  the  exponent  x.     Thus,  by  definition, 

e'  =  1  +  i  +  iV2  !  +  iV3  !  +  •••  +  i"/nl  +  ■■■. 

Series  for  a*.     Let  a  denote  any  positive  number  and  x  any     991 
real  number. 

Since  a  =  e^^Se",  §  732,  we  have  a^  =  e^'^-^",  §  730. 

Therefore,  substituting  x  log,  a  for  x  in  the  series,  §  990,  (2), 
we  have 
a*  =  1  -f  a:  log.a  +  x2(log,a)y2!  +  ■■■  +  x"{\og,ay/nl  -f  •  •  •. 

It  can  be  proved,  as  in  §  968,  1,  that  this  series  converges 
for  all  finite  values  of  x. 

The  logarithmic  series.     If  in  the  series  just  obtained  for  a^     992 
we  replace  a  by  1  -f  ic  and  x  by  y,  we  have 

(l+xy  =  l  +  \og,(l  +  x)-i/  +  l\og,(l  +  x)Yif/2l  +  .-..(l) 

But  by  the  binomial  theorem,  §  988,  when  |a;|  <  1, 

By  carrying  out  the  indicated  multiplications  and  collecting 
terms  we  can  transform  (2)  into  a  power  series  in  y. 
The  coefficient  of  y  in  this  series  will  be 

^  +  "i:^'^        1-2.3        +---'0^^       2  +  3  • 


558  A   COLLEGE   ALGEBRA 

Equate  this  to  the  coefficient  of  y  in  (1).     We  obtain,  if 

'  log,(l  +  ic)  =  a;-a;72  +  a;73-a;V4  +  -.-.  (3) 

This  series  is  called  the  lorjarlthmlc  series.  In  §  968  we 
proved  that  it  converges  when  |a;|  <  1. 

In  the  proof  just  given  we  have  assumed  that  the  series  (2)  remains 
equal  to  (1  +  x)^  after  it  has  been  transformed  into  a  power  series  in  y. 
But  this  follows  from  §976  when  la;|<l.  For  if  x'  and  y'  denote  |x| 
and  \y\  respectively,  the  series  Z7i'  +  Uo'  +  •  •  -of  §  976,  corresponding  to 

^  1-2  1.2.3 

and  this  series  is  convergent  when  x'  <  1,  its  sum  being  (1  —  x')-"',  §  988. 
As  we  liave  proved  the  truth  of  the  binomial  theorem  only  for  rational 
values  of  the  exponent,  it  may  be  observed  that  (3)  follows  from  (1)  and 
(2)  even  when  y  is  restricted  to  rational  values  (see  the  remark  at  the 
end  of  §972). 

Example.     Show  that  JV^^  {l  +  -\   =  e. 

We  have 

log/l  +  -y  =  nloge(l+-)=n(l--^ +  ...)  = 

Hence  1'™  log.f  1  +  ^V  =  1,  and  therefore    1™  fl  +  -  Y'  =  ei  =  e. 

For  limu  =  limeio-e"  =  e'"" dog,, «>,  §§  726-729,  73L 

993  Computation  of  natural  logarithms.  The  logarithms  of  num- 
bers to  tlie  base  e  are  caWed  their  7iafurallof/arith7)is.  A  table 
of  natural  logarithms  may  be  computed  as  follows : 

We  have  log, {l  +  x)  =  x-  xy2  +  x^S  -  x'/4  +  •  • .,  (1) 
and  therefore  log,  (l-x)  =  -x-  xy2  -  x^S  -  x*/4: .  (2) 

Subtract  (2)  from  (1).     Since 

1  +  iC 
log,(l  +a;)-log,(l  -  x)^\^g^YZr^' 


1  + 


1  +  a;  /         x^      x^  \ 

weobtain    loge  ^^T^  =  2  (^x  +  3  +  ^  +    ■■). 


(3) 


THE   LOGARITHMIC    SERIES  559 

I  A_  X      n  +  1                                           1 
In  (3)  set = and  therefore  x  = • 

We  obtain 

n+l_^f      1  1  1  \ 

^^'     n         *'\27i  +  l  ^3(2 w  +  1/^  5(271  + 1)^"^"7' 

or  log^(/i  +  1) 

=  ^°S^^^  +  ^  {jlhl  +  3  (2  n  +  1)3  +  5  (2  7.^  1)^  +  •  •  •)•  ^^^ 

Setting  ?i  =  1  in  (4), 

log,2  =  2(1/3  +  1/3*  +  1/5  .  35  +  •  •  •)  -  .6931  •  ■  •. 

Setting  ?t  =  2  in  (4), 

log,  3  =  log,  2  +  2  (1/5  +  1/3  •  5»  +  •  •  •)  =  1-0986  •  •  •, 
and  so  on  to  any  integral  value  of  71. 

Modulus.  By  §  755,  log„n  =  logg7i/log,a.  Hence  the  loga-  994 
rithnis  of  numbers  to  any  base  a  may  be  obtained  by  multi- 
plying their  natural  logarithms  by  l/log^a.  We  call  l/\og^a 
or  its  equivalent,  §  756,  log^e,  the  modulus  of  the  system  of 
logarithms  to  the  base  a.  In  particular,  the  modulus  of  the 
system  of  common  logarithms  is  logigg  =  .43429  •  •  •. 

EXERCISE   XC 

1.  Compute  loge4  and  logeS  each  to  the  fourth  decimal  figure. 

2.  Show  that  e-i  =  2/3  !  +  4/5  !  +  6/7  !  +  •  •  •. 

3.  Show  by  multiplication  that 

/ ,      X      x2      x3  \  / ,      X      x2      x3  \ 

('  +  I  +  2!  +  37-^---)('-i  +  2--3!+---)='- 

4.  Show  that  (e'^  +  e-  ^^)/1  =  1  -  xV2  !  +  xV-4  '•  -  xV6  !  +  •••. 

5.  Show  that  (e'^  -  e-  '^)/2  i  =  x  -  xV3 !  +  xV5  !  -  xV7  !  +  •  •  • 

6.  Show  that  the  (r  +  l)th  term  in  the  expansion  of  (1  —  x)-"  by  the 

.  1  ,,  .    n(n-fl)---(n  +  r-l)    ^ 

bmomial  theorem  is  — ^ ^ z^. 

r\ 

7.  Find  the  term  in  the  expansion  of  (8  +  x)^  which  involves  x*. 

8.  Find  the  term  in  the  expansion  of  (1  —  x^)~^  which  involves  x^. 


560  A    COLLEGE   ALGEBRA 

9.    For  what  values  of  x  will  the  expansion  of  (9-4a;2)i  and  of 
(12  +  z  +  X-)'  in  ascending  powers  of  x  converge? 

10.  Expand  (1  —  x  +  2  x2)Mn  powers  of  x  to  the  term  involving  x*. 

11.  Find  the  first  three  terms  of  the  expansion  of  (8  +  3  x)^  (9  —  2  x)~  ' 
in  powers  of  x.     For  what  values  of  x  will  the  expansion  converge  ? 

12.  Find  the  limiting  values  of  the  following  expressions. 

(1)  lim  ^'-^  -  ^'\  (2)  lim    (1  +  x^)^  -  (1  -  x'^)^ 

^"°        3x  ''^"°(H-3x)^-(l  +  4x)i" 


13.   Prove  that  '""("!+  -  V=  e^. 


14.  Prove  that  li«i  (e^  +  x)'  =  e^. 

15.  Expand  logc(l  +  x  +  x^)  in  powers  of  x  to  the  term  involving  x*. 
For  what  values  of  x  will  this  expansion  converge  ? 


/in  —  n\^      l/?/i  — n\8 


16.  Show  that  loge  —  = 

71  n  2 

17.  Show  that  loge  ^^  =  1  +  —  +  —  +  .. .. 

n2  -  1      n2      2  ?i*      3  n6 

18.  Show  that 

n  +  1      2(n  +  l)2      3(n  +  l)3  n      2n^      3n^ 


XXXV.     RECURRING   SERIES 

995         Recurring  series,     A    series  a^  +  a^x  +  a^x^  -\ in  which 

every  r  +  1  consecutive  coefficients  are  connected  by  an  identity 
of  the  form 

«n  +  i'lOn  -  1   +  i^2«„-2  "I h  i^/'n  -  r  =  0, 

where  ^1,  p2,  •  •  •,  j^r  ^^6  constant  for  all  values  of  n,  is  called  a 
recurring  series  of  the  xth  order,  and  the  identity  is  called  its 
scale  of  relation. 

Thus,  in         1  +  3x  +  5x2  +  7x3  +  .  •  •  +  (2n  +  l)x»  +  •  •  ■  (|) 

every  three  consecutive  coefficients  are  connected  by  the  formula 

a„-2a„_i  +  a„_2  =  0;  (2) 

for  5-2-3  +  1  =  0,  7-2.5  +  3  =  0,  2n  +  1- 2(2n -1)  +  2n  -  3  =  0. 


RECURRING    SERIES  561 

Hence  (1)  is  a  recurring  series  of  the  second  order  whose  scale  of 
relation  is  (2). 

A  geometric  series  is  a  recurring  series  of  the  first  order. 

Every  power  series  which  is  the  expansion  of  a  proper  frac- 
tion whose  denominator  is  of  the  rth  degree  is  a  recurring 
series  of  the  Ah  order. 

Thus,  if i^ — ^  =  ao+      aix  +      a^z^  + h  a„x»  +  •  •  ■ ,  (1) 


we  have  2  +  x  =  ao  +    ai  I  x  +    ao 

+  2ao|     +2ai 


X2  +  ---+      an 

+  2a„_i 
+  3n„_2 


+  3rto 

Therefore        «o  =  2,     ai  +  2  ao  =  1,     a™  +  2  ai  +  3  ao  =  0, 
and,  in  general,  o„  +  2a„_i  +  3a„_2  =  0.  (2) 

Hence  (1)  is  a  recurring  series  of  the  second  order  whose  scale  of  rela- 
tion is  (2). 

If  a  few  terms  at  the  beginning  of  a  power  series  be  given, 
it  is  always  possible  to  find  a  scale  of  relation  which  these 
terms  will  satisfy.  By  aid  of  this  scale  the  series  may  be 
continued,  as  a  recurring  series,  to  any  required  term. 

Example.  Given  1  -i-  4  x  -|-  7  x2  -i-  lOx^  -|- 13  x*  -|-  •  •  • .  Find  the  scale 
of  relation  which  these  terms  satisfy,  and  then  find  two  additional  terms. 

As  1  +  4  X  -1-  7  x2  -f-  •  •  •  is  not  a  geometric  series,  we  begin  by  testing  the 
scale  of  the  second  order,  a„  +  pan-i  +  qan-2  =  0- 

If  all  the  given  terms  are  to  satisfy  this  scale,  we  must  have 
1  +ip  +  q=:0,     10  +  'lp+4q  =  0,     13  +  10^  +  7^=0. 

Solving  the  first  two  of  these  equations,  p  =  —  2,  g  =  1. 

And  these  values  satisfy  the  third  equation,  since  13  —  10  •  2  +  7  =  0. 

Hence  the  required  scale  is  a„  —  2  a„_i  +  a„_2  =  0. 

We  may  therefore  find  the  required  coefficients  as,  a^  as  follows  : 
as  -  2  •  13  +  10  =  0,  .-.  as  =  16 ;     as  -  2  •  16  +  13  =  0,  .-.  ae  =  19. 

If  the  given  terms  had  been  1  +  4  x  +  7  x^  +  10  x^  +  14  x*,  they  would 
not  have  satisfied  a  scale  of  the  form  a„  +  pa»-i  +  qa„-2  =  0.  But  we 
might  then  have  assigned  a  sixth  term  ax^  arbitrarily  and  have  found  a 
scale  of  the  third  order,  a„  +pa„_i  +  ga„_2  +  ra,i-3  =  0,  which  the  six 
terms  would  satisfy,  namely,  by  solving  for  p,  q,  r  the  equations 
10  +  7p  +  4g  +  r  =  0,     14 +102)  +  7g +  4r  =  0,     a +14j) +105 +7r  =  0. 


562  A   COLLEGE   ALGEBRA 

Fiom  the  example  it  will  be  seen  that  ordinarily  when  2  r 
terms  are  given  the  series  may  be  continued  in  one  way  as  a 
recurring  series  of  the  rth  order,  and  that  when  2  r  +  1  terms 
are  given  it  may  be  continued  in  infinitely  many  ways  as  a 
recurring  series  of  the  {r  +  l)th  order. 

998  The  generating  function  of  a  recurring  power  series.  Every 
recurring  power  series  of  the  rth  order  is  the  expansion  of  a 
proper  fraction  whose  denominator  is  of  the  rth  degree  (com- 
pare §  996).  This  fraction  is  called  the  generating  function 
of  the  series.  It  is  the  sum  of  the  series  when  the  series  is 
convergent. 

Thus,  let  ao  +  aix  +  a-infl  + h  a„x"  +  •  ■  •  (1) 

be  a  recurring  series  of  the  second  order  ^'hose  scale  of  relation  is 

««  +  i'ctn-i  +  ga„-2  =  0.  (2) 

Set        (S„=ao+    aix+    aoX^H +      a„_ix"-i 

.•.px<S„=         jiaoX  +  paix^H 1-    j)a„_2X"-i  +  pa„_iX'' 

.■.qx^Sn=  qaiyX^-\ h    ga„  _3X»-i+ q'a„_2X"  +  (7a„_iX''+i 


.:{l  +  px  +  qx'^)Sn  =  ao+{ai  +  pao)x  +(pa„_i  +  ga„_2)a;"        +qan-ix"+^ 

the  remaining  terms  on  the  right  disappearing  because  of  (2). 

When  (1)  is  convergent,  as  n  increases  iS„  will  approach  S,  the  sum 
of  (1),  as  limit,  and  x"  will  approach  0. 

Therefore  {1  +  px  +  qx"^)  S  =  a^  +  (ai  +  pao)  x, 

that  is,  S  =  "o  +  (^i+?>«")^  (3) 

1  +  px  +  gx2  ^  ' 

999  The  general  term  of  a  recurring  power  series.  This  may  be 
obtained,  Avhen  the  generating  function  is  known,  by  the 
method  illustrated  in  the  following  example. 

Example.  Find  the  generating  function  and  the  general  term  of  the 
recurring  series  whose  scale  is  a„  —  a,i-i  —  2  a„_2  =  0  and  whose  first  two 
terras  are  6  +  4  x. 

Here  j)  =  — 1,     g  =  _  2,     oq  =  5,     ai  =  4. 

Therefore,  by  §  998,  (3),  S  =  — ^^^ —  = ^-^^ -.  (1) 

'     -^  ^         ^  "  1-X-2X2       (l  +  x)(l-2x)  ^  ' 


RECURRING   SERIES  663 


Separating  (1)  into  its  partial  fractions,  §  537, 
b-  X  2 


+ =  2  (1  +  X)- 1  +  3 (1  -  2  x)-i. 


(l  +  x)(l-2x)       1+x       1-2X 

But  if   lx|<l,         2(l  +  x)-'  =  2[l-x  +  x2 +  (_i)«x« +  ••■]• 

And  if  I X I  <  1/2,  3  (1  -  2  X)- 1  =  3  [1  +  2  X  +  4  x2  +  ■ .  •  +  2"x''  +  •••].. 
Therefore  the  general  term  is  [(—  1)"2  +  3  •  2"]x". 

EXERCISE  XCI 

1.  If  the  first  three  terms  of  a  recurring  series  of  the  third  order  are 
2  -  3  X  +  5  x2  and  the  scale  of  relation  is  a„  +  2  a„_i  —  a„_2  +  3  otn-s  ==  0, 
find  the  fourth  and  fifth  terms. 

2.  Find  the  scale  of  relation  and  two  additional  terms  in  each  of  the 
following: 

(1)  1  +  3  X  +  2  x2  -  x3  -  3  X*  +  •  •  • . 

(2)  2  -  5  X  +  4  x2  +  7  x'  -  26  X*  +  •  •  ■ . 

(3)  1 -3x +  0x2 -10x3  +  15x4-21x6  + ■•  •• 

3.  Find  the  generating  function  and  the  general  term  of  each  of  the 

following : 

(1)  2 +  x  +  5x2  + 7x3  + I7x*  +  •••. 

(2)  3  +  7x  +  17x2  +  43x3  +  113x*  +  --.. 

4.  Prove  that  in  a  recurring  series  Uq  +  UiX  +  •  •  •  of  the  third  order, 

whose  scale  of  relation  is  a„  +  pa„^i  +  qa„-2  +  rans  =  0,  the  generating 

.    cto  +  («!  +  pao)  x  +  (a2  +  pai  +  qao)  x'^ 

function  IS  — ^—; ; 

1  -\-  px  +  gx2  +  rx3 

5.  By  aid  of  the  preceding  formula  find  the  generating  function  and 
the  general  term  of  the  series 

1  +  2X  + 11x2 +  24x3 +  85x4 +  238x5+  .... 

6.  Show  that  a  +  (a  +  (i!)  x  +  (a  +  2  d)  x2  +  (a  +  3  d)  x^  +  . .  •  is  a  recur- 
ring series  of  the  second  order,  and  find  its  generating  function. 

7.  Show  that  12  +  22x  +  32x2  +  42x3  +  •  •  •  is  a  recurring  series  of  the 
third  order  whose  generating  function  is  (1  +  x)/(l  —  x)3. 

8.  Show  that  l-2  +  2-3x  +  3-4x2  +  4-5x3  +  -..  is  a  recurring 
series  of  the  third  order,  and  find  its  sum  when  convergent. 


564  A    COLLEGE    ALGEBRA 

XXXVL     INFINITE    PRODUCTS 

1000  Infinite  products.  This  name  is  given  to  expressions  of  the 
form 

n  (1  +  a,)  =  (1  +  «i)  (1  +  «2)  •  ■  •  (1  +  «>)  ■■■: 

in  which  the  number  of  the  factors  is  supposed  to  be  infinite. 
Such  a  product  is  said  to  be  convergeyit  or  divergent  accord- 
ing  as    (1  +  fli)  (1  +  "2)  •••(!  +  a„)    approaches    or  does  not 
approach  a  finite  limit  as  n  is  indefinitely  increased. 

1001  Theorem.  If  all  the  nuvibers  a^  are  positive,  the  infinite 
product  n  (1  +  a^)  is  convergent  or  divergent  according  as  the 
infinite  series  Sa,.  is  convergent  or  divergent. 

First,  suppose  that  2a^  is  convergent  and  has  the  sum  S. 
Then  since  1  +  a;  <  e^  when  x  is  positive,  §  990, 

we  have  (1  +  a-^)  (1  +  r/o)  •  •  •  (1  +  a,,)  <  e"!  ■  e"^  ■  ■  ■  e"-, 
that  is,  <  e"i  +  "2  +  - ••  +  «'.<  e"?. 

Hence,  as  n  increases,  (1  +  a^)  (1  +  a„)  •  ■  -(1  +  «„)  increases 
but  remains  less  than  the  finite  number  e^.  It  therefore 
approaches  a  limit,  §  192,  that  is,  11(1  +  0^)  is  convergent. 

Second,  suppose  that  1a^  is  divergent. 

In  this  case  lim  (a^  +  ^2  +  •  •  •  +  "„)  =  ^• 

But  (1  +  «i)  (1  +  «2)  ••■(!  +  fl„)  >  1  +  (fli  +  oo  +  •  •  •  +  a„). 

Therefore  lim  (1  +  Oj)  (1  +  Un)  ■  ■  ■  (1  +  o,,)  =  00,  that  is, 
n  (1  +  a,.)   is   divergent. 

Thus,  n  (1  +  1/n'')  is  convergent  when  j)  >  1,  divergent  when  p  <  1. 

Example  1.  If  2ar  is  a  divergent  positive  series  whose  terms  are  all 
less  than  1,  show  that  n  (1  —  a^)  =  0. 

Since  a^  <  1  and  1  —  af.  <  1,  we  have  1  —  a,.  <  1/(1  +  a,)  numerically. 
Hence  (1  -  a{)  (1  -  aa)  •  •  •  (1  -  a,)  <  1/(1  +  a^)  (1  +  a.)  ■  ■  •  (1  +  a„). 
But  lim  (1  +  ai)  (1  +  a.)  •  •  ■  (1  +  a„)  =  00. 

Therefore   n  (1  -  a,)  =  lim  (1  -  ai)  (1  -  a^)  ■  ■  ■  (1  -  a„)  =  0. 


INFINITE    PRODUCTS  565 

Example  2.     Show  that  when  x  =  1  the  binomial  series,  §  987,  con- 
verges if  jn  +  1  >  0,  but  diverges  if  m  +  1  <  0. 
When  X  =  1,  the  binomial  series  becomes 

l  +  m  +  '^^i^+...+    -         -'-        -■-'+•••.        (1) 


In  this  series  we  have 


+  • 

,  m{m-l)---(m-n  +  l)    , 

'                 1.2. .-n                  ' 

m  - 

- n  +  1  _             m  +  1 

(2) 


Hence,  if  m  +  1  <  0,  we  have  |  w„  + 1/«„  |  >  1,  and  (1)  is  divergent,  §  951. 

But  if  m  +  1  >  0,  after  a  certain  term  the  series  will  be  of  the  kind 
described  in  §  958  and  will  therefore  converge. 

For  if  r  denote  the  first  integer  greater  than  ?h  +  1,  it  follows  from  (2) 
that  when  n>r,  u„  +  i/u„  is  negative  and  numerically  less  than  1.  Hence 
the  terms  of  (1)  after  Ur  are  alternately  positive  and  negative  and  decrease 
numerically.     Therefore  (1)  converges  if  lim  w„  =  0. 

„  in    m  —  1        m  —  n  +  1 

But    u„  +  i  =  -.^ 

12  n 

=<-'K'-^)('-'^>-(-'^)- 

and  it  follows  from  Ex.  1  that  as  n  increases  the  product  on  the  right 
approaches  0  as  limit. 

EXERCISE  XCII 

,     ^^         ^      3    5    9    17  .5    10    17    26 

1.  Show  that  -  —  -.-...  and  -  •  —  —  —  .  •  •  are  convergent. 

2    4    8    16  4     9     16    25 

2.  For  what  positive  values  of  x  are  the  following  infinite  products 
convergent  ? 

<^>  "('<i:)=('-p)o+l)('4:)- 
<^>  "0-s)K'-f,)('^i:)o-i-;)-- 
<'>  "0-i-:)=o-i)(>+f)('-s>- 

3.  Show  that  u^„'^(«  +  ^)(«  +  2)---(«  +  n)^  ^  ^^  ^  according  as 

^^  b{b  +  l)(b  +  2)...{b+n) 

o  >  0  or  a<o. 


566  A    COLLEGE    ALGEBRA 


XXXVIL     CONTINUED    FRACTIONS 

1002  Continued  fractions.     This  name  is  given  to  expressions  of  the 

form  a+-        ,           ,  or  «  +  -  ,   -  ,         as  they  are  usually 
c  +d  G-\-  e  -\ 

e  +  --- 
written. 

We  shall  consider  simple  continued  fractions  only.  These 
have  the  form  a^^ —   ,        ,  where  a^  is  a  positive  integer 

«o  +  «3  +      ■;  ,^ 

or  0  and  «,>  «3)  •  •  •  ^^^  positive  integers. 

The  numbers  a^,  a.^,  •■•  are  called  the  first,  second,  •  ■■partial 
quotients  of  the  continued  fraction. 

According  as  the  number  of  these  quotients  is  finite  or  infi- 
nite, the  fraction  is  called  terminating  or  nonterminating. 

1003  Terminating  fractions.  Evidently  every  terminating  simple 
continued  fraction  has  a  positive  rational  value,  for  it  may  be 
reduced  to  a  simple  fraction.. 

1       1      ^       4        .30       1       1       1       13 

Thus,  2  +  -^^  =  2  +  -  =  ~;     2  +  3  +  4  =  35- 

Conversely,  every  positive  rational  number  may  be  converted 
into  a  terminating  simple  continued  fraction.  This  will  be 
evident  from  the  following  example. 

Example.     Convert  67/29  into  a  continued  fraction. 
Applying  the  method  for  finding  the  greatest  common  divisor  of  two 
integers  to  67  and  29,  we  have 
29)67(2  =  ai 
58 
9)  29  (3  =  aa 
27 
2)  9  (4  =  as 
8 

1)2(2  =  04 
2 
0 


^      2+  '^  =2+     '    . 
29              29              29/9 

(1) 

!=-h-^- 

(2) 

^-i 

(3) 

CONTIXUED    FRACTIONS  567      ' 

Substituting  (2)  in  (1),  and  (3)  in  the  result,  we  have 

67      „       1  „       1       1       1 

■ —  =  2  +  -  =2  +  -       -       -,as  required. 

29  3  +  1  3  +  4  +  2  ^ 

4+1 

2 

Since  29/67  =  1  h-  67/29,  we  also  have 

29_  1       1       1       1 

67  "2  +  3  +  4  +  2* 

Convergents.     The   fractio^^       .->  Oi+— ?  «i  +  —      — '    •••,  1004 
1  a^  «2  +  <^z 

are  called  the  first,  second,  third,  •  •  •  convergents  of  the  fraction 
11 

oi  +  —       — 

«2   +   '^'3  H Q 

When  Oj  is  0,  the  first  convergent  is  written  — • 

Theorem  1.    Each  odd  convergent  is  less  and  each  even  con-  1005 
vergent  is  greater  than  every  subsequent  convergent. 

This  follows  from  the  fact  that  a  fraction  decreases  when 
its  denominator  increases. 

Thus, 

1.  ai<ai  +  •  •  •. 

11  .11 

2.  ai  H >ai  -\ ,  since  —  >  — 

a^  a2-\ a2     02  +  •  •  • 

1  1  .11 

3.  ai  ^ <ai-\ )  since  a^ -{ —  >(H-\ » 

1  „     ,    1  «3  as  +  --- 

02-1 02   H 

Os  Os  +  •  •  • 

by  2;  and  so  on. 

Reduction  of  convergents.    On  reducing  the  first,  second,  third,  1006 

•  •  •  convergents  of  aj  H —  to  the   form  of   simple 

.        .  ,     .  «2  +  *s  +  •  •  • 

tractions,  we  obtain 

«i     (i\(^2  +  1     ayi„^nj^r_a^j\-_a^ 
1  ftj  ''a^s  +  1  ' 


568  A    COLLEGE   ALGEBRA 

Let  2>i,  i?2)  V^i  ■  ■  ■  denote  the  numerators,  and  g-j,  g-j,  g'g,  •  •  • 
the  denominators  of  the  convergents  as  thus  reduced,  so  that 

IH  =  «i,  IH  =  «i«2  +  Ij  i^s  =  aia2«3  +  «!  +  «3,  •  •  •         (2) 
^1  =  1,     q2  =  a^,  3-3  =  a2«3  + 1,  ••••  (3) 

Since  «!,  a.^,  a^,  ■  ■  •  are  positive  integers,  it  follows  from  (2), 
(3)  that  as  n  increases  p^  and  q^  continually  increase,  and 
that  they  approach  oo  if  the  given  fraction  does  not  terminate. 

By  examining  (2)  and  (3)  it  will  be  found  that 

2h  =  «3i>2  +  Pi   and   q^  =  a^q^  +  q^.  (4) 

This  is  an  illustration  of  the  following  theorem. 

1007  Theorem  2.  The  numerator  and  denominator  of  any  conver- 
gent are  connected  with  those  of  the  two  preceding  convergents  by 
the  formulas 

Pn  =  a^Pn-l  +  Pn-2,        qn  =  a„q„  _  i  +  q„_2- 

For  suppose  that  these  formulas  have  been  proved  to  hold 
good  for  the  A;th  convergent,  so  that 

Vk  =  (^kPk  - 1  +  Pk-2,     qk  =  %?i-  -i  +  qk-i,  (1) 

Pk  ^  (^kPk-l   +Pk-2  ^2) 

qk     (ikqk-i  +  qk-2 

The  (k  +  l)th  convergent  may  be  derived  from  the  kth.  by 
merely  replacing  a^  by  a^.  +  l/a,^_^.^,  §  1004.  Therefore,  since 
Pk-v  Pk-if  qk-ij  qk-2  do  not  involve  a^.,  it  follows  from  (2)  that 

Pk  +  i  _(ak  +  'i~/ak  +  -i)Pk-i  +Pk-2 

qk + 1     («x-  +  i/«i- + 1)  qk  -i  +  qk-2 

^  ak+i{fi'kPk-x  +  Pk-2)+Pk-x  ^  ak  +  ^Pk  +  Pk-i    by(l)^ 
that  is, 

pk+i  =  ^h-^iVk+Pk-v    qk+i  =--  "k+iqk  +  qk-v 

We  have  thus  proved  that  if  the  formulas^,,  =  a„p„_i  +jo„_j> 
q^  =  a„q„_i  +  q„_2  hold  good  for   any  particular  convergent, 


CONTINUED    FRACTIONS  569 

tliey  hold  good  for  the  next  convergent  also.  But  we  have 
already  shown  that  they  hold  good  for  the  third  convergent. 
Hence  they  hold  good  for  the  fourth,  hence  for  the  fifth,  and 
so  on  to  every  convergent  after  the  thh-d  (compare  §  791). 

Example.     Compute  the  convereeiits  of  3  +  - 

^  ^  ^  2+3+4+5 


Since   3  =  3/1   and   3  +  1/2  =  7/2,    we   have  pi  =  3, 
Hence  P3  =  3  •  7  +  3  =  24,  p4  =  4  •  24  +  7  =  103,  ps  =  5  • 

P2  =  7,  gi  =  l, 

103  +  24  =  539, 

and           53  =  3  •  2  +  1  =    7,  54  =  4  •    7  +  2  =:    30,  55  =  5  • 

•    30+    7  =  157. 

^,       ,         ,                                  3     7     24     103     539 

Therefore  the  convergeuts  are  -,  -,  — ,  , 

1     2      7       30      157 

Theorem  3.      The  numerators  and  denominators  of  every  two   1008 
consecutive  converyents  are  connected  hy  the  formula 

Puqa-l-Pa-iqn=(-l)°- 

The  formula  holds  good  when  n  —  2.  For,  by  §  1006,  we 
have  /»2?i  —  P\<1'2.  =  («i«2  +  1)  —  «'i«2  =  1  =  (—  1)^- 

Moreover  we  can  prove  that  if  the  formula  holds  good  when 
n  =  k,  it  also  holds  good  when  71  —  k  -{-  1. 

For  2Jk +i1k-  JMk + 1  =  ((h  +  i2h  +  ]h - 1)  'Ik  - Pk («i-  +  iqk  +  <lk - 1) 
=  -{Pk<h-^-Pk-xqk)-  §1007 

Hence,  if  p^q^  _  1  -  7^,.  _  1  fh  =  (-  l)^ 

tl^en  p^  +  iqk-~  Pklk  + 1  =  (-  1)"^^  + '. 

Therefore,  since  the  formula  is  true  for  n  =  2,  it  is  true  for 
«,  =  2  +  1  or  3,  therefore  for  7i  =  3  +  1  or  4,  and  so  on  to  any 
positive  integral  value  of  n. 

Corollary  1.    Every  convergent  ^J(\^  is  an  irreducible  fraction.  1009 

For  if /»„  and  §'„  had  a  common  factor,  it  would  follow  from 
the  relation  2'„7„_  1 —jt)„_i2'„  =(— 1)"  that  this  factor  is  a 
divisor  of  (—  1)",  which  is  impossible. 


570  A    COLLEGE    ALGEBRA 

1010  Corollary  2.  For  the  differences  between  conver gents  we  have, 
the  forntidas 

1        Pn  Pn-1_(-1)°  ^        Pn  Pn- 2  ^   ("  1)° "  ^a„ 

qn  Cln-l  qnqn-1  "'      qn  qn-2  qnqn  -  2 

The  first  formula  is  an  immediate  consequence  of  the  x<s\2^- 
tion;;„^„_i-j;„  _,./„=(-  1)". 

The  second  follows  from  the  fact,  §§  1007,  1008,  that 

Pnin  -2-Pn-2<ln='  {(^nPn  -l+Pn-2)  Qn  -  2  "  l^n  -  2  («„?«  -  1  +  2'n  -  2) 

=  ««(i>«-i?«-2-i\-29'„-i)  =  (-  !)""'«„• 
The  theorem  of  §  1005  may  be  derived  from  these  formulas. 

1011  Theorem  4.  The  nth  convergent  of  a  no7iterminating  simple 
continued  fraction  approaches  a  definite  limit  as  n  is  indefinitely 
increased. 

For,  by  §  1005,  the  odd  convergents  pi/qi,  Pz/q^,  •  •  •  form  a 
never-ending  increasing  sequence,  every  term  of  which  is  less 
than  the  finite  number  p^/q^.  Hence,  §  192,  a  variable  which 
runs  through  this  sequence  will  increase  toward  some  number 
A  as  limit. 

Similarly  a  variable  which  runs  through  the  sequence  of 
even  convergents  p^/q^,  Pi/qa  •  •  •  will  decrease  toward  some 
number  /a  as  limit. 

But/A  =  X,since/x-A=  lim  f^^-^^i^!!^"]  =,  lim  (-1)""^^^ 

"'^  "  L'^ '»         ^2  m  - 1 J         "'=  -  q-Z  ,„q-2  ,n  - 1 

Therefore  a  variable  which  runs  through  the  complete 
sequence  of  convergents  Pi/qi,P2/qi,  Ih/qz,  •••  will  approach 
X  as  limit. 

1012  By  the  value  of  a  nonterminating  simple  continued  fraction 
is  meant  the  number  li^ii  (pjq^).  It  follows  from  §  1003  that 
this  number  is  always  irrational. 

The  value  of  a  terminating  fraction  is  that  of  its  last  con- 
vergent, §  1004. 


COXTIXUED   FRACTIOXS  571 

In  the  statements  of  the  following  theorems,  §§  1013,  1014, 
the  understanding  is  that  when  the  fraction  terminates,  "  con- 
vergent'"' means  any  convergent  except  the  last  one. 

Corollary  1.     The  value  of  a  simple   continued  fraction  lies   1013 
hetwmn  the  values  of  every  two  consecutive  convergents. 

Corollary  2.     The  difference  betiveen  the  value  of  a  continued  1014 
fraction  and  that  of  its  nth  co7iverr/ent  is  numerically  less  than 
l/^n^ln  +  i  ^'^'^  greater  than  a„  +  ,/qnq„^2- 

For  let  X  denote  the  value  of  the  fraction,  and  to  fix  the 
ideas  sujipose  that  n  is  odd. 


We  then  hav( 
Hence 
and 

3 

X- 
X 

In           In +  2                        'In  +  l 
Pn^J'n  +  l          P",       •      ^           1          , 

1005,  1013 
§  1010,  1 
§  1010,  2 

?„           'lu  +  r           9n                       'Jn'In  +  l 
'/«           'In +  2          'In                       1n'In  +  2 

Evidently  l/q^qn  +  i  <  ^/'I%  ^^^^  ^J  leaking  use  of  the  rela- 
tion «7„_^2  =  ^'«  +  2'Z«+i  +  'Inj  §  1007,  it  may  readily  be  shown  that 
•^» +  2/7n'7n  + 2  =  !/'/«('?«  + I7n  +  i)-  Hcncc  the  difference  between 
X  and  2Jn/'Li  is  less  than  1/y^  and  greater  than  l/'/„(q„  +  q„  +  i)- 

Corollary  3.     JSach  convergent  is  a  closer  approximation  to  the   1015 
value  of  the  fraction  than  is  any  preceding  convergent. 

For,  by  §  1014,  if  X  denote  the  value  of  the  fraction,  the  dif- 
ference between  X  dcnfi  p ,J q ,^  is  numerically  less  than  '^/q„qn  +  \, 
while  the  difference  between  A.  and  p„_  ,/2'„_i  is  numerically 
greater  than  o„  +  ,/./„  _  ^q„  + 1 ;  and  l/^?,//,,  + 1  <  «« +  i/q„ _ ,7,,  + 1, 
since  q„-i<  <'n  +  iqn^  §  1006. 

Corollary  4.     The  convergent  Pn/cin  ^^  "'  <^loser  approximation    1016 
to  the  value  of  the  fraction  than  Is  any  other  ratloiial  fraction 
whose  denominator  does  not  exceed  q„. 

For  if  a/b  is  a  closer  approximation  to  the  value  of  the 
fraction   than  Pn/qn   i^?    ^^   must    also,    §  1015,    be    a   closer 


572  A   COLLEGE    ALGEBRA 

approximation  than  7>„_i/j„_i  is,  and.  must  therefore,  §1013, 
lie  between  ])„lqn  ^^^  Pn  -  il'L,  - 1- 

To  fix  the  ideas,  suppose  that  n  is  even. 


We  then 

have 

qn-i     ^     qn 

Hence 

qn 

that  is, 

1       ^  "V«-i-^7>.-. 

qnq.  - 1  "q,.  - 1 

or  i>>  q„{'iqn-i-k\-\)- 

But  a'7„_i  —  ^>Pn-i  is  positive,  since  a/h  >  ^>„  _  i/q„_  ^  Hence 
b  >  q„;  that  is,  if  a/b  is  a  closer  approximation  to  the  value  of 
the  continued  fraction  than  p,Jq„  is,  its  denominator  b  must  be 
fj renter  than  //„. 
1017  Recurring  fractions.  A  nonterminating  continued  fraction 
in  which  a  single  partial  quotient  or  a  group  of  consecutive 
partial  quotients  continually  recurs  is  called  a  recurrlmj  frac- 
tion. And  such  a  fraction  is  called  pure  or  mixed  according 
as  it  begins  or  does  not  begin  with  these  recurring  partial 
quotients. 

The  value  of  a  recurring  fraction  may  be  found  as  follows. 

Example  1.     Find  the  value  of2  +  -  =2  +  -      -      - 

3  +  --.  3  +  3  +  3  +  --. 

This  is  a  pure  recurring  fraction  with  ihe  period  2  -\ Hence,  if  x 

denote  its  value,  we  have  '"' 

x:.2  +  l      1,    .•.x=I^+^,  .•.3x^-Gx-2  =  0,  .■.x  =  ^±^. 
3  +  X  3x  +  l  3 

Example  2.     Find  the  value  of  4  +  - 

5  +  2  +  3  +  .  • . 

This  is  a  mixed  recurring  fraction  with  the  period  2  +  1/3.     Hence,  if 
X  denote  the  value  of  the  recurring  part  2  +  ^  ,  and  y  tlie  value  of 

the  entire  fraction,  we  have,  by  Ex.  1, 

1       1       21 X  +  4       21  (3  +  vT5)/3  +  4        75  +  21  VTs 


y  =  4 


5  +  x       5x  +  l         5(3  +  Vl6)/3  +  l  18  +  5  VTs 


CONTINUED    FRACTIONS  573 

In  general,  if  x  denote  the  value  of  a  pure  recurring  fra.ction 

with  the  period.  «i  +  •  •  •     — '  we  have,  §  1007, 

+  ('k 

x  =  a,  +  ...     1       l^P^+JP^, 
+  a^  +  x       q„x+qf,_^ 

and  therefore      qf.x^  +  (5-^,  _  1  —  2h)  *'  ~  2h  _  1  =  0. 

Since  the  absolute  term  —  p<.  _i  of  this  quadratic  is  negative, 
it  has  one  and  but  one  positive  root,  and  this  root  is  the  value 
of  the  fraction. 

Again,  if  1/  denote  the  value  of  the  mixed  recurring  fraction 

i       J_  J_ 

+  a,  +  a,  +1  H +  a,._^i.  H 

we  find  the  value  x  of  the  recurring  part  as  above,  and  then 
have,  §  1007, 

1         1        ]->,.x  +  J)r-1 

«  =  «,  +  •••     —      -=  — - — — • 

+  a^  +  x       q,:x  +  qr-i 

On    converting    irrational    numbers    into    continued   fractions.    1018 

Every  positive  irrational  number  is  the  value  of  a  definite 
nonterminating  simple  continued  fraction  which  may  be 
obtained  to  any  required  partial  quotient  by  the  following 
process. 

If  h  denote  the  number  in  question,  first  find  a^,  the  greatest 
integer  less  than  h.  Then  b  =  Oi  +  l/^i,  where  h^  is  some  irra- 
tional number  greater  than  1.  Next  find  (u,  the  greatest  integer 
less  than  b^.  Then  b^  =  a^  +  l/b^,  where  b^  is  some  irrational 
number  greater  than  1.     Continuing  thus,  we  have 

&  =  oi  +  —  =  rtj  H —  =  ...  =  «j  H — 

^^1  "o  +  b^  02  +  «3  H 

It  can  be  proved  that  when  5  is  a  quadratic  surd  the  con- 
tinued fraction  thus  obtained  is  a  recurring  fraction. 


574  A   COLLEGE   ALGEBRA 

Example.     Convert  Vu  into  a  continued  fraction. 
The  greatest  integer  less  tlian  VTl  is  3,  and,  §  60.3, 

Vll  =  3  +  (Vrr-3)  =  3  +  — J =3  +  -— J (1) 

_Vll  +  3  (Vll  +  3)/2 

The  greatest  integer  less  than  (Vu  +  3)/2  is  3,  and 

^  ^  2  ( Vn  +  3)  Vu  +  3 

The  greatest  integer  less  than  vTl  +  3  is  G,  and 

Vri  +  3  =  G  +  (VIl -3)  =  6  + ^ =6  + ? (3) 

Vll  +  3  (VTi  +  3)/2 

The  last  fraction  in  (3)  is  the  same  as  the  last  in  (1).  Hence  the  steps 
from  (3)  on  will  be  (2),  (3)  repeated  indefinitely  ;  that  is,  the  partial  quo- 
tients 3  and  G  will  recur.     Hence,  sub.stituting  (2)  in  (1),  and  (3)  in  the 

result,  and  so  on,  we  have  Vll  =  3  +  - 

3  +  G  +  ... 

1019  A  given  irrational  number  can  be  expressed  in  onli/  one  tray 
as  a  simple  continued  fraction.  This  follows  from  the  fact 
that  two  nonterminating  simple  continued  fractions  cannot  be 
equal  unless  their  corresponding  partial  quotients  are  equal. 

For  if  a  +  or  =  c  +  7,  where  a  and  c  denote  positive  integers  and  a 
and  7  denote  positive  numbers  which  are  less  than  1,  then  a  =  c,  since 
otherwise  it  would  follow  from  a  —  c  =  y  —  a  that  an  integer,  not  0,  is 
numerically  less  than  1. 

Hence,  if  aiH —  =  Ci -I —  ,  where  ai,  a2,  as,  ••  •, 

"2  +  "3  H C2  +  C3  H 

...  ^  11 

Ci,  C2,  Cs,  •  •  •  denote  positive  integers,  we  have  Oi  =  Ci,  .-.  —      — 

a2  +  «3  +  •  •  • 
11  11 

=  —      —  , .-.  tto  H —  =  ("2  H —  ,  .-.  02  =  C2,  and  so  on. 

C2  +  Cs  H '         03  H Cs  +  ■  ■  • 

1020  If  we  compute  the  continued  fraction  to  which  a  given 
irrational  number  b  is  equal  as  far  as  the  nth  partial  quotient, 
we  can  find  its  nth  convergent  p„/q„,  and  this  rational  fraction 
p„/q„  will  express  b  approximately  with  an  error  less  than 
1/ql,  §  1014.  Moreover  p„/qr.  will  be  a  closer  approximation 
to  b  than  is  any  other  rational  fraction  whose  denominator 
does  not  exceed  q„,  §  1016. 


CONTINUED    FRACTIONS  575 

Thus,    the   first   four   convergents   of  vll  =  34--      -      _  are 

3  +  G  +  3  +  ... 

3    10    68    100         ^   100  /—     .^^  ,       ^,         1 

- ,  —  ,  —  — ,  and  expresses  Vll  with  an  error  less  than 

1     3     19     GO  GO  60'^ 

Solution  of  indeterminate  equations  of  the  first  degree.  Given  1021 
any  equation  of  the  form  ax  +  hij  =  c,  where  a,  b,  c  denote 
integers  of  which  a  and  h  have  no  common  factor,  §  672,  If 
we  convert  a/b  into  a  continued  fraction,  the  last  convergent 
of  this  fraction  will  be  a/b  itself,  and  if  the  convergent  next 
to  the  last  be  p/q,  we  have  aq  —  bjy  =  ±  1,  §  1008.  This  fact 
makes  it  possible  always  to  find  a  jjair  of  integral  values  of  x 
and  //  which  satisfy  ax  +  bi/  =  c.  The  method  is  illustrated 
in  the  following  example. 

Example.     Find  an  integral  solution  of  205  x  +  93  y  =  7. 

Asin  §1003,  Ex.,  wefind  —  =2  +  i      111. 

93  4  +  1  +  8  +  2 

2     0     11     97     205 

The  convergents,  found  as  in  §  1007,  Ex.,  are  ",  -,  — ,  — ,  ^• 

^        '  ^  '         '         1     4      5      44      93 

Hence  205  •  44  -  93  ■  97  =  -  1, 

or,  multiplying  by  -  7,   205  (-  44  •  7)  +  93  (97  •  7)  =  7. 

Therefore  x  =  -  308,  y  =  679  is  a  solution  of  205  x  +  93  y  =  7. 

The  general  solution  is  x  =  -  308  +  93 1,  y  =  679  -  205 1,  §  674. 

Similarly  we  may  show  that  205  x  —  03  y  =  7  has  the  solution  x  =  —  308, 
y  =  -  679. 

EXERCISE  XCIII 

Compute  the  convergents  of  the  following : 

1.3  +  1   1   '.  2.  1  ^-   1   ±   1. 

4  +  1  +  5  1  +  1+3  +  10  +  12 

Convert  each  of  the  following  into  a  continued  fraction.  For  each  of 
the  last  three  compute  the  fourth  convergent  and  estimate  the  error  made 
in  taking  this  convergent  as  the  value  of  the  fraction. 

3.   '±  4.    1^.  5.    11?.  6.    3.54. 

12  56  513 

7.    .1457.  8.   ?^.    .     ,         9.   ^.  10.   '^. 

177     >  ^  972  31827 


576  A    COLLEGE    ALGEBRA 

Convert  each  of  the  following  into  a  recurring  continued  fraction  and 
compute  the  fifth  convergents  and  the  corresponding  errors  for  the  first 
four  of  them. 

11.    Vl7.  12.    V2G.  13.    V(3.  14.    VsS. 

15.    Vl05.  16.    I/V23.  17.    VlO.  18.    VtT. 

19.    3  Vs.  20.    (Vl0-2)/2.         21.    (V2  +  l)/( V2  -  1), 

Find  the  values  of  the  following  recurring  fractions. 

22.    i      1      i  .     23.    i      1       i  .     24.    34-^      i     i  • 

1  +  2+3  +  ---  2  +  1  +  3  +  ---  4  +  5  +  2+    •• 

25.    ^  +  1      ^       i  .  26.    1      1      i      ^      i  . 

3  +  4  +  5  +  ...  2  +  7  +  1  +  2  +  1  +  --. 

27.    Show  that  Va^  +  1 


29.  Show  that 

ill  _  -  {abc  +  a-b  +  c)  +  V(abc  +  a  +  6  +  c)^  +  4 

a  +  b  +  c  +  ---~  2  (a6  +  1) 

30.  Convert  the  positive  root  of  x-  +  x  -  1  =  0  into  a  continaed 
fraction. 

31.  Show  that  ^-^  =  ^  +  -i L  +  .-.  +  lnil:'. 

Qn     3i     qiq2     (Mz  qn-iqn 

11  111 

32.  Show  that  -       —  = 1 . 

a-i  +  as  -] 91^/2      5293      qsqi 

33.  What  rational  fraction  having  a  denominator  less  than  1000  will 
most  nearly  express  the  ratio  of  the  diagonal  of  a  square  to  its  side  ? 
Estimate  the  error  made  in  taking  this  fraction  as  the  value  of  the  ratio. 

34.  Find  the  simplest  fraction  which  will  express  tt  =  3.14150265-  •• 
■with  an  error  which  is  less  than  .000001. 

35.  Compute  the  sixth  convergent  of  e  =  2.71828  •  •  ■  and  estimate  the 
error  made  in  taking  it  as  the  value  of  e. 

36.  Find  an  integral  solution  of  127 x  -2Uy  =  6. 

37.  Find  an  integral  solution  of  235  x  +  412  y  -  10. 

38.  Find  the  general  integral  soluticJh  of  517  x  -  323  j/  =  81. 


PROPERTIES   OE   CONTINUOUS   EUNCTIONS      577 

XXXVIII.     PROPERTIES    OF   CONTINUOUS 
FUNCTIONS 

FUNCTIONS   OF  A  SINGLE  VARIABLE 

Functions.     If  the  variable  y  depends  on  the  variable  x  in  1022 
such  a  manner  that  to  each  value  of  x  there  corresponds  a 
definite  value  or  set  of  values  of  y,  we  call  y  a.  function  of  x. 

In  what  follows  when  we  say  that  ?/  is  a  function  of  x  and 
write  y  =f(x),  we  shall  mean  that  it  is  a  one-valued  function; 
in  other  words,  that  to  each  value  of  x  there  corresponds  but 
one  value  of  //.  And  f{a)  will  denote  the  value  of  y  which 
corresponds  to  the  value  a  of  x. 

Evidently  ?/  is  a  function  of  x  if  it  be  equal  to  an  algebraic  expression 
in  X,  as  when  y  =  x^  +  1.  But  a  relation  which  defines  y  as  a  function 
of  X  may  be  one  which  cannot  be  expressed  by  an  equation.  Thus,  y  is 
a  function  of  x  if  y  is  1  for  all  rational  values  of  x  and  —  1  for  all  other 
values  of  x.  But  this  relation  between  y  and  x  cannot  be  expressed  by  an 
equation. 

We  call  y  a  function  of  x  even  when  there  are  exceptional  values  of  x 
foi  which  the  given  relation  between  y  and  x  fails  to  determine  y,  §  1024. 

•Sometimes  y  is  defined  as  a  function  of  x  only  for  a  certain  class  of 
values  of  x  or  only  for  values  of  x  which  lie  between  certain  limits. 
Thus,  the  equation  ?/  =  x  +  2x2  +  3x''  +  •  ■  •,  by  itself  considered,  deter- 
mines y  for  those  values  only  of  x  which  are  numerically  less  than  1. 

Continuity  of  a  function.     Let  f{x)  denote  a  given  function  1023 
of  X.    We  say  that  f{x)  is  continuous  at  a,  that  is,  when  x  =  a, 
if /(«)  has  a  definite  finite  value,  and  if  ^^"^  f{x)=f{a). 

In  the  contrary  case  we  say  that /(a?)  is  discontinuoiis  at  a. 

Here  and  in  what  follows  the  notation  ^^"^  f(x)  =  f(a)  means 
that  f(x)  will  approach  /(a)  as  limit  whenever  x  approaches 
a  as  limit,  that  is,  no  matter  what  the  sequence  of  values  may 
be  through  which  x  runs  in  approaching  a  as  limit. 

In  the  case  of  a  function  y  defined  by  a  given  equation  1024 
y=f(x)  it  may  happen  that  the  expression /(x)  assumes  an 


578  A   COLLEGE    ALGEBRA 

indeterminate  form  when  x  =  a,  §§  513-518.  The  equation 
y  =f(^x)  by  itself  considered  then  fails  to  define  ?/  when  x  =  a. 
But  if  ^^^^  f(x)  has  a  definite  finite  value,  we  assign  this  as  the 
value  of  /(«),  §  519,  which  makes  f(x)  continuous  at  a.  If 
li^ /(a;)  =  00,  we  assign  to  /(«)  the  value  oo,  §  515;  f(x)  is 
then  discontinuous  at  a.  Finally,  if  ^^^^  /(^)  is  indetermi- 
nate, we  have  no  reason  for  assigning  any  single  value  to 
f{o-).  Evidently  we  can  assign  none  for  which  ^i^^  f(^)  =f{^)- 
In  this  case  also  f(x)  is  discontimwus  at  a. 

1.  Thus,  every  rational  function/(a;)  is  continuous  except  perhaps  when 
the  denominator  of  some  fraction  occurring  in/(x)  vanishes. 

For  example,  consider  the  function  /(x)  =  (x  —  l)/(x^  —  1). 

This  function  is  continuous  except  when  x"^  —  \  =  0,  that  is,  when 
a;  =  1  or  —  1.  For  if  a  is  not  1  or  —  1,  /(a)  =  (a  —  l)/(a'-  —  1)  has  a 
definite  finite  value  and  ''i'^^/(x)  ==/(a),  §  509. 

When  X  =  1,  the  expression  (x  —  Vj/{x^  —  1)  assumes  the  indeterminate 
form  0/0.  But  l™/(a;)  =  ^"^^  [(^  -  l)/(x-^  -  1)]  =  ^'i"  [l/(a;  +  1)]  =  1/2, 
and  by  assigning  to  /(I)  the  value  1/2  we  make /(x)  continuous  when 
x  =  \. 

When  X  =  —  1,  /(x)  is  discontinuous  ;  for   ^l"^  /(x)  =  oo. 

2.  Consider  the  following  function  : 

/(x)=4^=ii^==i^^. 

2^+1       1  +  1/2^       1/2   ^-  +  1 
Here/(0)  has  the  indeterminate  form  oo/oo,  §  517. 
But  if  we  write  /(x)  in  the  second  form  and  tlien  make  x  approach  0 

through  positive  values,  we  have  lim  2-^  =  oo,  and  therefore  lim/(x)  =  1. 

If  we  write  f(x)  in  the  third  form  and  then  make  x  approach  0  through 
_  1 
negative  values,  we  have  lim  2    *  =  co,  and  therefore  lim/(x)  =  .3. 

Finally,  if  we  make  x  approach  0  through  values  which  are  alternately 
positive  and  negative,  /(x)  will  not  approach  any  limit. 

Hence /(x)  is  discontinuous  at  0.  No  value  can  be  assigned  to/(0)  for 
which  I'm /(x)=/(0).  -' 

1025        From  the  definition  of  continuity  in  §  1023  it  immediately 
follows,  §  189,  that 
V 


1026 


PROPERTIES   OF    CONTINUOUS   FUNCTIONS      579 

The  sufficient  and  necessary  condition  that  f  (x)  he  coritimwiis 
at  a  is  that  f  (a)  have  a  definite  finite  value,  and  that  for  every 
positive  number  S  which  can  be  assigned  it  shall  be  possible  to 
find  a  corresponding  positive  miniber  e  such  that 

|f  (x)  —  f  (a)|  <  8  whenever  [x  —  a[  <  e. 

Thus  in  the  neighborhood  of  a  value  of  x,  as  a,  at  which 
f{x)  is  continuous,  very  small  changes  in  the  value  of  x  are 
accompanied  by  very  small  changes  in  the  value  of  f{x), 
and  the  change  in  the  value  of  x  can  be  taken  small  enough 
to  make  the  corresponding  change  in  the  value  oi  f{x)  as  small 
as  we  please.  This  is  not  true  of  f{x)  in  an  interval  con- 
taining a  value  of  x  at  which  f(^x)  is  discontinuous.  See  the 
examples  in  §  1024. 

Theorem  1.  If  both  of  the  functions  f  (x)  and  ^(x)  are  con- 
tinuous at  a,  the  same  is  true  of  f(x)±<^(x)  and  f(x)-^(x)j 
also  of  f  (x)/<^  (x)  unless  ^  (a)  =  0. 

Ifi{^)  is  continuous  at  a,  the  same  is  true  of  vf  (x). 

This  follows  immediately  from  the  definition  of  continuity 
at  a,  §  1023,  and  the  theorems  of  §§  203-205,  according  to 
which  lim  \f{x)  +  ^(.r)]  =  \\m.f(x)  +  lim  ^{x),  and  so  on. 

Real  functions.    In  what  follows  x  will  denote  a  real  variable,  1027 
that  is,  one  which  takes  real  values  only,  and  f(x)  will  denote 
a  real  function  of  x,  that  is,  one  which  has  real  values  when  x 
is  real. 

Number  intervals.     The  practice  of  picturing  real  numbers  by  1028 
points  on  a  straight  line,  §§  134,  209,  suggests  the  following 
convenient  nomenclature. 

Let  us  call  the  assemblage  of  all  real  numbers  between  a 
and  b,  a  and  b  themselves  included,  the  number  interval  a,  b, 
and  represent  it  by  the  symbol  (a,  b). 

Moreover,  it  being  understood  that  a  <b,  let  us  call  a  and 
b  the  left  and  right  extremities  of  the  interval  (a,  b).     Also,  if 


/■ 


580  A   COLLEGE   ALGEBRA 

c  =  (a  +  ^)/2,  let  us  say  that  (a,  h)  is  divided  at  c  into  the  two 
equal  intervals  (a,  c)  and  (c,  b)]  and  so  on. 

Thus,  (1,  7)  is  divided  at  4  into  the  two  equal  intervals  (1,  4),  (4,  7); 
and  at  3  and  5  into  the  three  equal  intervals  (1,  3),  (3,  5),  (5,  7). 

1029  We  say  that  the  function  f{x)  is  continuous  throughout  the 
interval  (a,  b)  if  it  is  continuous  for  every  value  of  x  in  this 
interval. 

1030  Theorem  2.  Ifi  (x)  is  contimious  throughout  the  interval  (a,  b), 
and  f  (a)  and  f  (b)  have  contrary  signs,  there  is  in  (a,  b)  a  number 
Xq  such  that  f  (Xq)  =  0. 

To  fix  the  ideas,  suppose  that /(a)  is  +  and  that/(^)  is  — . 
Divide  {a,  b)  into  any  number  of  equal  intervals,  say  into  the 
two  equal  intervals  (a,  c)  and  (c,  b). 

If /(c)  =0,  our  theorem  is  proved,  c  being  Xq.  But  if/('")  =?^  0, 
it  must  be  true  of  one  of  the  intervals  (a,  c)  or  (c,  b)  that/(.r) 
is  +  at  its  left  extremity  and  —  at  its  right.  Thus,  if  /(r) 
is  — ,  this  is  true  of  {a,  c),  and  iif(c)  is  +,  it  is  true  of  (c,  b). 
Select  this  interval  and  for  convenience  call  it  (a^,  bi).  Then 
/(«i)is  +  and/(^*i)is  -• 

Deal  with  this  interval  («!,  b^)  as  we  have  just  dealt  with 
(a,  b),  and  so  on  indefinitely.  We  shall  either  ultimately  come 
upon  an  interval  extremity  for  which  f(x)  =  0,  which  is  then 
the  Xq  sought,  or  we  shall  define  a  never-ending  sequence  of 
intervals  within  intervals, 

(a,  b),  («!,  Z*i),   (a^,  b.^,   •••,  (a„,  b„),  •••, 

such  that         f{a),  f(a{),  /(a.,),  •  •  • ,  f(a„),  ■■■  are  + 

and  /(/.),  f(b,),  f(b,),  . .  .,f(b,,),    ...  are  -. 

It  follows  from  §§  192,  193  that  as  n  increases  a„  and  b,^ 
approach  the  same  number  as  limit.     For  a„  remains  less  than 
b  and  never  decreases,  and  b,^  remains  greater  than  a  and  never 
increases,  and  lim  (^>„  —  a,,)  =  lim  (b  —  «)/2"  =  0. 
Thenf(xo)=0. 


PROPERTIES   OF    CONTINUOUS   FUNCTIONS      581 

For  since/(£c)  is  continuous  at  x„,  lim/(a„)  =  lim/(6„)  =/(xo). 

But  since /(«„)  is  always  positive,  its  limit /(xq)  cannot  be 
negative,  and  since  /(^„)  is  always  negative,  its  limit  /(x^) 
cannot  be  positive.     Therefore  /(x^)  is  0. 

Thus,  if  f{x)  =  1  -  xV2  !  +  a;V4  !  -  x^/6  !  +  •••,  it  may  readily  be 
shown  that  /(I)  is  positive  and  /(2)  negative.  Hence  this  f(x)  will 
vanish  for  some  value  of  x  between  1  and  2. 

Simpler  illustrations  of  the  theorem  will  be  found  in  §§  833,  836. 

Maximum  and  minimum  values.     Superior  and  inferior  limits.   1031 

Consider  the  following  infinite  assemblages  of  numbers : 

2,H,  li,li,  ...(A),         2,  2i,2|,2|,  ...(B). 

In  (A)  there  is  a  greatest  number,  namely  2,  but  no  least 
number ;  and  in  (B)  there  is  a  least  number,  namely  2,  but 
no  greatest  number. 

On  the  other  hand,  while  there  is  no  least  number  in  (A), 
among  the  numbers  which  are  less  than  those  in  (A)  there  is  a 
greatest,  namely  1.  Similarly  among  the  numbers  which  are 
greater  than  those  in  (B)  there  is  a  least,  namely  3. 

The  like  is  true  of  all  infinite  assemblages  of  finite  numbers, 
that  is,  of  numbers  which  lie  between  two  given  finite  numbers 
a  and  b.     In  other  words, 

Theorem  3.     Let  aj,   a^,   •••,  a„,  ••.   (A)  denote  any  infiniie  1032 
assemblage  of  finite  numbers.     Then 

1.  Either  among  the  different  numbers  in  (A)  there  is  a 
greatest  or  among  the  numbers  greater  than  those  in  (A)  there 
is  a  least. 

2.  Either  among  the  different  numbers  in  (A)  there  is  a  least 
or  amo7ig  the  mimbers  less  than  those  i^i  (A)  there  is  a  greatest. 

To  prove  1  assign  all  numbers  greater  than  those  in  (A)  to 
a  class  7?2)  and  all  other  real  numbers,  including  those  in  (A), 
to  a  class  Ri.  Since  each  number  in  7?i  will  then  be  less  than 
every  number  in  i?.,-  there  will  be,  §  159,  either  a  greatest  num- 
ber in  i?i  or  a  least  in  R2,  —  which  means  either  a  greatest 


582  A   COLLEGE   ALGEBRA 

among  the  different  numbers  in  (A)  or  a  least  among  the  num- 
bers which  are  greater  than  those  in  (A). 
By  similar  reasoning  2  may  be  proved. 

1033  If  among  the  different  numbers  of  an  assemblage  there  is 
a  greatest,  we  call  that  number  the  maximum  number  of  the 
assemblage ;  if  a  least,  its  minimum  number. 

The  superior  limit  of  an  assemblage  is  the  maximum  number, 
if  there  be  one.  If  not,  it  is  the  least  number  which  is  greater 
than  every  number  in  the  assemblage. 

The  inferior  limit  of  an  assemblage  is  the  minimum  number^ 
if  there  be  one.  If  not,  it  is  the  greatest  number  which  is  lesa 
than  every  number  in  the  assemblage. 

An  assemblage  like  1,  2,  3,  4,  •  •  •  which  contains  numbers 
greater  than  every  assignable  number  is  said  to  have  the  supe- 
rior limit  00.  Similarly  an  assemblage  like  —1,  —2,  —3, 
—  4,  •  •  •  is  said  to  have  the  inferior  limit  —  co . 

Evidently,  if  an  assemblage  has  a  finite  superior  limit  X, 
either  A.  is  its  maximum  number  or  we  can  find  in  the  assem- 
blage numbers  which  differ  from  X  as  little  as  we  please. 

1034  By  the  "  values  of /(a-)  in  (a,  h)  "  we  shall  mean  those  which 
correspond  to  values  of  x  in  (a,  h).  And  if  this  assemblage 
has  a  maximum  or  a  minimum  value,  we  shall  call  it  the  abso- 
lute maxim/um  or  minimum,  value  of  f{x)  in  (a,  li).  The  maxi- 
mum and  minimum  values  defined  in  §  639  may  or  may  not 
be  the  absolute  maximum  and  minimum  values. 

1035  Theorem  4.  If  f  (x)  is  continuous  thronghout  the  interval 
(a,  b),  it  has  an  absolute  maximum  and  an  absolute  minimum 
value  in  (a,  b). 

For  since  the  values  of /(a-)  in  (a,  b)  are  finite,  §  1023,  they 
have  finite  superior  and  inferior  limits.  Call  these  limits  A 
and  /A  respectively. 

We  are  to  demonstrate  that  in  (a,  b)  there  is  a  number 
Xq  such  that  f{x^  —  k,  and  a  number  x^  such  that  f{x{)  =  fi. 


PROPERTIES   OF    CONTINUOUS    FUNCTIONS      583 

As  the  proofs  of  these  two  theorems  are  essentially  the  same, 
we  shall  give  only  the  first  of  them. 

Divide  (a,  b)  into  any  number  of  equal  intervals,  say  into 
two  such  intervals.  Evidently  A.  will  be  the  superior  limit 
of  the  values  of  f(x)  in  at  least  one  of  these  half  intervals. 
For  convenience  call  this  half  interval  («!,  b^). 

Deal  with  the  interval  («i,  ^i)  as  we  have  just  dealt  with 
(a,  b),  and  so  on  indefinitely. 

We  thus  obtain  a  never-ending  sequence  of  intervals  within 
intervals, 

(a,b),  {a„b,),  (a,,b,),   ■■.,   («„,  ^.„),   .••, 

in   each    of    which   A    is    the    superior    limit    of    the    values 
of/(x). 

As  71  is  indefinitely  increased,  «„  and  i„  approach  the  same 
number  as  limit  (see  §  1030). 

If  we  call  this  limit  x^,,  then  /(a"o)  =  X. 

For  if  not,  since  both  /(x^)  and  X  denote  constants,  their 
difference  must  be  some  constant,  as  a,  different  from  0, 
so  that 

X-f(^,)=a.  (1) 

Since  f(x)  is  continuous  at  x^,  we  can  make  the  interval 
(a„,  b„)  so  small  that  for  every  value  of  x  in  (a„,  6„)  we  have, 
§  1025, 

\f(x)-f{x,)\<a/2.  (2) 

And  since  X  is  the  superior  limit  of  the  values  of  f(x)  in 
(a„,  b„),  we  can  choose  x  in  (a„,  ^>„)  and  (2)  so  that,  §  1033, 

X  -f(x)  <  a/2.  (3) 

But  it  will  then  follow  from  (2)  and  (3)  that 

X  -/(xo)  <  a.  (4) 

Therefore,  since  (4)  contradicts  (1),  (1)  is  false ;  that  is, 
\  —f(xo)  =  0,  or  A  =f(Xo),  as  was  to  be  proved. 


584  A    COLLEGE   ALGEBRA 

1036  Corollary.  T/f  (x)  is  continuous  throughout  the  interval  (a,  b), 
it  will  have  in  (a,  b)  every  value  intermediate  to  its  maximum 
and  miyiimxim  values  in  this  interval. 

For  let  c  denote  the  value  in  question  and  consider  the 
function  f(x)  —  c,  which  is  continuous  in  (a,  b),  §  1026. 

If/(.ro)  and/(a-i)  denote  the  absolute  maximum  and  mini- 
mum values  of  /(.r)  in  (a,  b),  /(r„)  —  c  is  +  and  /(xj)  —  c  is  — . 
Hence,  §  1030,  between  x^  and  a-i  there  is  a  number,  call  it  x^, 
such  that/(j:-.,)  —  c  =  0,  or/(a-2)  =  c,  as  was  to  be  proved. 

1037  Oscillation  of  a  function.  By  the  oscillation  of  f(x)  in  (a,  b) 
is  meant  the  difference  between  the  superior  and  inferior  limits 
of  the  values  of /(.r)  in  (a,  b). 

1038  Theorem  5.  Let  f(x)  he  continuous  tliroughout  (a,  b).  If 
any  positive  number  a  be  assigned,  hoicever  small,  it  is  possible 
to  divide  (a,  b)  into  a  finite  number  of  equal  intervals  in  each  of 
which  the  oscillation  off(x)  is  less  than  a. 

For  divide  (a,  b)  into  any  number  of  equal  intervals,  say 
into  two  such  intervals,  each  of  these  in  turn  into  two  equal 
intervals,  and  so  on.  The  process  must  ultimately  yield  inter- 
vals in  each  of  which  the  oscillation  of /(.«)  is  less  than  a. 

For  if  not,  there  must  be  in  {a,  b)  at  least  one  half  interval 
in  which  the  oscillation  of  f{x)  is  not  less  than  a ;  in  this,  in 
turn,  a  half  interval  in  which  the  oscillation  of  f(x)  is  not  less 
than  a ;  and  so  on  without  end. 

Let  this  never-ending  sequence  of  intervals  within  inter- 
vals be 

i<',h),   (a„b,),   (a,,b,),   ...,   (a„,b„),   .■■, 

and,  as  in  §  1030,  let  lim  a„  =  lim  b„  =  .r„. 

Since  f(x)  is  continuous  throughout  (a,  h),  it  has  an  abso- 
lute maximum  and  an  absolute  minimum  value  in  each  of  the 
intervals  (a,  b),  (a^,  bi),  •••,  (a„,  b„),  •••,  §  1035. 

Let  /(«-„)  denote  the  absolute  maximum  and  f(^„)  the 
absolute  minimum  value  of  f(x)  in  (a„,  b„). 


PROPERTIES   OF    CONTINUOUS   FUNCTIONS      585 

Then,  by  hypothesis,         /K)-/(A.)>«. 
and  therefore  lim  f{a^)  —  lim  /(/?„)  >  a. 

But  this  is  impossible.  For  since  a„  and  /?„  are  in  (a„,  &„), 
and  lim  a„  =  lim  h„  =  x^,  we  have  lim  «„  =  lim  ^„  =  x^. 

Therefore,  since /(a-)  is  continuous  at  x^,  lim/(a„)  =  lim/(^„); 
that  is,  lim/(a„)  —  lim/(/3„)  =  0,  .'.  not  >  a. 

FUNCTIONS   OF   TWO   INDEPENDENT  VARIABLES 

Functions  of  two  variables.     We  say  that  the  variable  u  is  a  1039 
function  of  the  variables  x  and  y  when  to  each  pair  of  values  of 
X  and  ?/ there  corresponds  a  definite  value  or  set  of  values  of  u. 

We  shall  confine  ourselves  to  the  case  in  which  to  each  pair 
of  values  of  x,  y  there  corresponds  a  sijigle  value  of  u. 

The  notation  u  —f(x,  y)  will  mean  that  w  is  a  function  of 
x  and  y,  and  f(a,  h)  will  mean  the  value  which  u  has  when 
x  =  a  and  y  —  h. 

Thus,  M  is  a  function  of  x  and  y  if  u  =/(x,  y)  =  x"^  -  2y  -\-\.  Here, 
when  X  =  1,  ?/  =  2,  we  have  u  =/(!,  2)  =  1-4  +  1=  —  2. 

The  note  at  the  end  of  §  1022  applies,  mutatis  mutandis,  here  also. 

Continuity  of  such  a  function.     Let  f(x,  y)  denote  a  given  1040 
function  of  x  and  y.     AVe  say  that /(a;,  y)  is  continuous  at  a,  b, 
that  is,  when  x  =  a  and  y  =  b,  if  f(a,  b)  has  a  definite  finite 
value  and  \if(x,  y)  will  always  approach /(rt,  b)  as  limit  when 
X  and  y  are  made  to  approach  a  and  b  respectively  as  limits. 

In  the  contrary  case  we  say  that  f{x,  y)  is  discontiruwns  at 
a,  b,  that  is,  when  x  =  a  and  y  =  b. 

From  this  definition  and  §  189,  it  immediately  follows  that 

The  sufficient  and  necessary  conditio7i  that  f  (x,  y)  be  continu-   1041 
ous  at  a,  b  is  that  f  (a,  b)  have  a  definite  finite  value,  and  that  for 
every  j^ositive  number  S  which  can  be  assigned  it  shall  be  possible 
to  find  a  corresponding  positive  number  e  such  that 

|f  (x,  y)  —  f  (a,  b)|  <  8  whenever  |x  —  al<  e  and  [y  —  b|  <  c 


586 


A   COLLEGE    ALGEBRA 


Theorem  1.  If  both  of  the  functions  f  (x,  y)  and  ^  (x,  y)  are 
continuous  at  a,  b,  tlte  same  Is  true  of  i(x,  y)±  <f>(x,  y)  and 
f  (X,  y)  •  <^(x,  y),  also  ofi(x,  y)/</>(x,  y),  u?iless  <^(a,b)=0. 

Ifi(x,  y)  Is  co7itlmious  at  a,  b,  the  same  is  true  of  Vf  (x,  y). 

This  follows  immediately  from  §  1040  and  §§  203-205. 

Number  regions.  In  what  follows  it  is  to  be  xmderstood  that 
X  and  7/  denote  real  variables,  and  /(.«,  y)  a  real  function  of 
these  variables  (compare  §  1027). 

As  is  shown  in  §  382,  pairs  of  values  of  x  and  ?/  may  be 
pictured  by  points  in  a  plane.  Evidently,  if  employing  this 
method  we  draw  the  lines  wh.  :h  are  the  graphs  of  the  equa- 
tions X  =  a,  X  =  b,  y  =  c,  7/  =  d,  i  384,  the  rectangle  bounded 
by  these  lines  will  contain  the  graphs  cf  all  pairs  of  values  of 
X,  y  such  that  a<x<b,  c<y<d.  With  this  rectangle  in  mind, 
we  shall  call  the  assemblage  of  all  such  pairs  of  values  of  x,  y 
the  number  region  (a,  b;  c,  d). 

We  say  that  f{x,  y)  is  contlmious  thronghoxit  the  region 
(a,  b ;  c,  d)  if  it  is  continuous  for  every  pair  of  values  of  x,  y 
in  this  region. 

Theorem  2.  //"f  (x,  y)  be  continuous  tli ronghout  the  reglo7i(?i,hi 
c,  d),  it  has  a  maximum  and  a  minimum  value  in  this  region. 

Since  f{x,  y)  is  continuous  throughout  the  given  region,  its 
values  within  this   region  have  finite   superior  and  inferior 
limits,  §§  1032,  1040.    Call  these 
limits  A.  and  fx.. 

We  are  to  prove  that  in  (a,  b ; 
f,  d)  there  is  a  value  pair  .To,  ?/„ 
such  that /(.To,  yo)=^;  and  simi- 
lar reasoning  will  show  that  there 
is  also  a  value  pair  x^,  y^  such 
that/(xi,  ?/i)  =  /i. 
^'  For    construct    tlie    rectangle 

EFGH  which  pictures  the  number  region  (a,  b ;  c,  d),  §  1043. 


l 

^ 

a 

77            77, 

G\^ 

I'\ 

F 

Ci 

E 

X 

0      ( 

(, 

<i 

b 

PROPERTIES   OF   CONTINUOUS    FUNCTIONS      587 

By  the  "  values  of /(.r,  y)  in  EFGH"  we  shall  mean  the  values 
of  f(x,  y)  corresponding  to  all  pairs  of  values  of  x,  y  in 
(a,  b  ;  c,  d). 

Divide  EFGH  into  four  equal  rectangles  as  in  the  figure. 
Evidently  A  will  be  the  superior  limit  of  the  values  of  f(x,  y) 
in  at  least  one  of  these  quarter  rectangles.  Call  this  quarter 
rectangle  EiF^GiH^. 

Deal  with  the  rectangle  EiF^GyH^  as  we  have  just  dealt 
with  EFGH,  and  so  on  indefinitely.  We  thus  obtain  a  never- 
ending  sequence  of  rectangles  within  rectangles, 

EFGH,  E.F^GJI^,  ■■■,  E^F^G^H^,  .■•,  (1) 

in  each  of  which  X  is   the   superior  limit   of  the  values  of 

f(^,  y)- 

Let  a„  denote  the  abscissa  of  ^„,  and  c„  its  ordinate.  As  is 
proved  in  §  1030,  when  n  is  indefinitely  increased  a„  and  c„ 
approach  definite  limits. 

If  lim  a„  =  Xq  and  lim  c^  ~  y^,  then  /(Xq,  y^  =  X. 

For  if  not,  let  X  -f(xo,  y^)  =  a.  (2) 

Since  f{x,  y)  is  continuous  at  x^,  y^,  we  can  so  choose 
E„F„G„H„  that  for  every  pair  of  values  of  x,  y  in  this  rec- 
tangle we  have,  §1041, 

\f(x,y)-f(x„y,)\<a/2.  (3) 

And  since  X  is  the  superior  limit  of  the  values  of /(a-,  y)  in 
E„F„G„H„,  we  can  so  choose  x,  y  in  E^FJJ^H^  and  in  (3)  that 

X-f{x,y)<a/2.  (4) 

From  (3)  and  (4)  it  then  follows  that 

^  -  fi^o,  2/0)  <  «•  (5) 

But  (5)  contradicts  (2).  Hence  (2)  is  false  and  therefore 
/{xq,  2/0)  =  X,  as  was  to  be  demonstrated. 


588  A  COLLEGE  ALGEBRA 

THE  FUNDAMENTAL  THEOREM  OF  ALGEBRA 

We  are  now  in  a  position  to  prove  that  evenj  rational  inte- 
gral equation  has  a  root,  §  797.     We  proceed  as  follows. 

1046  Theorem.      Given  (^  (z)  =  1  +  bz'"  +  cz'"  +  ^  -^ \-  kz",  ivhere 

b,  c,  •••,  k  denote  constants,  real  or  complex,  and  z  a  complex 
variable;  it  is  always  j)ossible  so  to  choose  z  that  |^(z)|<  1. 

For  let  the  expressions  for  z  and  h  in  terms  of  absolute 
value  and  amplitude,  §  877,  be 

z  =  p (cos  6  +  i  sin  6),  h  =  \b\-  (cos  yS  +  i  sin  /8). 

Then  ^<s;'"=p"'I^^|- [cos(m^  +  y8)  + isin(m^  +  y8)].  §§879,881 

First  choose  6  so  that  mO  +  ft  =  tt.  (1) 

Then      bz"'  =  p"'[^'|- (cos  tt  +  i  sin  7r)  =  —  p"'\b\, 

since  cos  tt  =  —  1  and  sin  tt  =  0.  §§  877,  878 

Next  choose  p  so  that,  §  854, 

|,|p™  +  i  +  ...+|7,lp»<jS|p'"<l.  (2) 

If  Zq  denote  the  value  of  z  which  corresponds  to  the  values 
of  6  and  p  thus  chosen,  then  |<^(«o)  1  <  1- 

For  since    <^ (^o)  =  (1  -  p'"\b\)  +  cz"^  +  '  +  ■  ■  ■  +  kz^, 
we  have,  §  235, 

\<^{z,:)\<\-p-\b\-Y\c\p-^'^----\-\k\p\  .•.<!,  by  (2). 

1047  Corollary.    Given  the  function  f  (z)  =  a„z"  +  aiz"  - '  H h  a„  ; 

if  f  (z)  does  not  vanish  when  z  =  b,  vie  can  always  choose  z  so 
that\i{z)\<\i(y)\. 

For  in  f{z)  set  z  =  b  +  h  and  develop  by  Taylor's  theorem, 
§848.  It  may  happen  that  certain  of  the  derivatives  f'(z), 
f"{z),  and  so  on  vanish  when  z  =  b;  but  they  cannot  all  vanish 


PROPERTIES   OF   CONTINUOUS   FUNCTIONS      589 

since  /^"X-)  =  w !  ^o-     I^et  /'"(s;)  denote  the  first  one  which  does 
not  vanish. 

Then  f{b  +  h)  =f(b)  ■\-r{h)  ^  +  ■  •  •  +/«(^)  ^, 

and  therefore      \.,,,       =  1  +  *'-— ^7  " ~f  "• +^,;,    •  — :• 

Jib)  Jib)     m\  fib)     n\ 

The  second  member  of  the  last  equation  is  a  polynomial  in 
h  of  the  form  considered  in  §  1046.  Hence  we  can  so  choose 
h  that  \f{b  +  h)/J\b)\  <  1  and  therefore  \f{b  +  f')\<W)\- 

Theorem.     Given  f  (z)  =  a„z"  +  aiZ°  ~  ^  +  •  •  •  +  a„  ;  a  value  of  1048 
z  exists  for  which  f  (z)  vanishes. 

For  in  fiz)  set  z  =  x  +  iy,  where  x  and  y  are  real,  and 
having  expanded  aQ[x  +  iy)",  a^i^x  +  iy)"~^,  •  •  •  by  aid  of  the 
binomial  theorem,  collect  all  the  real  terms  in  the  result,  and 
likewise  all  the  imaginary  terms.  We  may  thus  reduce /(«) 
to  the  form  f(z)  =  <^  (x,  y)  +  iij/  (x,  ?/),  where  (f>  (x,  y)  and  i/^  (x,  y) 
denote  real  polynomials  in  x,  y,  and  therefore  have,  §  232, 

\f{z)\  =  ^<^i^x,yy  +  ^{x,yyi^\ 

By  §  855,  we  can  find  a  positive  number,  as  c,  such  that  the 
roots  of /(,?)—  0,  if  there  be  any,  are  all  of  them  numerically 
less  than  c  ;  and  if  c'  =  c/^2,  evidently  |,t|,  or  (a-^  +  t/^)^,  is  less 
than  c  for  all  values  of  x,  y  such  that  —  c'  <x<c\  —  c'  <y<  c'. 

But  in  this  number  region  (—  c',  c' ;  —  c',  c')  the  expression 
[(f>(x,  y)'^  +  i(/(x,  ?/)^]^  is  a  continuous  function  of  x  and  y, 
§  1042.  It  therefore  has  a  minimum  value  in  this  region, 
§  1045,  say  when  x  =  Xq,  y  =  y^. 

If  z,  =  x,+  iy„  then  1/(^0)1  =  [<^(^o,  l/oY  +  ^(xo,  yo)l'  =  0. 

For  since  |/(«o)|  is  the  minimum  value  of  |/(s;)],  we  cannot 
make  \f(z)\  <  \f(zo) \.  Therefore  \f(zo) \  =  0,  since  otherwise, 
§1047,  we  could  so  choose  z  thai  \f(z)\<\f(zo)\. 

Hence  |/(«)|,  and  therefore  f(z),  vanishes  when  z  =  z^; 
that  is,  Zg  is  a  root  of  the  equation  /(s)  =  0. 


INDEX 


Numbers  refer  to  pages 


Abscissa,  138 

Addition  of  integral  expressions,  93 

of  numbers,  10, 19, 35,  50,71,  72 

of  radicals,  274 

of  rational  expressions,  217 

of  series,  541 
Amplitude  of  complex  number,  488 
Angle,  circular  measure  of,  488 
Annuities,  391 
Approximations,  48,  55,  453 
Assemblage,  infinite,  3 
Associative  law  of  addition,  11,  22, 
35,  54,  74,  521 

of   multiplication,  14,  23,  35, 
54,  74 
Asymptote,  335 

Base  of  power,  39 

of  system  of  logarithms,  377 
Binomial  theorem,  256,  283,  554 
Binomials,  products  of,  102,  253 
Biquadratics,  112,  48G 

Cardan's  formula  for  cubic,  483 

Chance,  409 

Clearing  of  fractions,  118,  231 

Coefficient,  86 

Coefficients,  detached,  99 

undetermined,  method  of,  152 
undetermined,  theorem  of,  172, 
540 

Cologarithms,  386 

Combinations,  393 


Commensurable,  37 
Commutative  law  of  addition,  11, 
22,  35,  54,  74,  534,  544 

of  multiplication,  14,  23, 35,  54, 
74 
Completing  the  square,  187,  300 
Condition,  necessary,  sufficient,  93 
Constants,  79 
Continuity  of  functions,  577,  585 

of  real  system,  46 
Convergence  of  infinite  series,  520 

absolute  and  conditional,  533 

limits  of,  536 

tests  of,  523,  531 
Convergents  of  a  continued  frac- 
tion, 567 
Converse,  92 
Coordinates,  138 
Correspondence,  one-to-one,  1 
Cosine,  489 
Counting,  9 
Cube  root.     See  Roots 
Cubics,  112,  483 

irreducible  case  of,  485,  490 
Cyclo-symmetry,  248 

Degree  of  equation,  111 

of  polynomial,  87 

of  product,  98 
Density  of  rational  system,  34 

of  real  system,  46 
Derivatives,  460 
Descartes's  rule  of  signs,  447 


591 


592 


A   COLLEGE   ALGEBRA 


Determinant,  494 

bordering  a,  505 

cofactors  of,  504 

diagonals  of,  496 

elements  of,  494 

evaluation  of,  505 

minors  of,  502 

order  of,  495 

products  of,  506 

properties  of,  498 

terms  of,  496 
Differences,  method  of,  364 
Discriminant,  517 

of  cubic,  485 

of  quadratic,  304 
Distributive  law,  14,  23,  35,  54,  74 
Divergence  of  infinite  series,  520 
Divisibility,  exact,  28,  155,  161 
Division  of  integral  expressions,  107 

of  numbers,  27,  35,  54,  73,  489 

of  radicals,  287 

of  rational  expressions,  219 

of  series,  546 
Division,  synthetic,  166 
Division  transformation,  155 

by  aid  of  undetermined  coeffi- 
cients, 160,  163 

Elimination,  131,  143,  317 

by  determinants,  508,  514 
Ellipse,  334 
Equality,  3,  8,  32,  34,  45,  72 

algebraic  and  numerical,  18.  '^5 

rules  of,  13,  15,  24,  36,  54,  57 
Equations,  binomial,  313,  490 

biquadratic,  112,  486 

complete,  426,  448 

conditional,  110 

cubic,  112,  483 

depressed,  427 

equivalent,  117,  131 

exponential,  390 


Equations,  fractional,  111,  231, 
300 

identical,  89 

inconsistent,  133,  146 

indeterminate,  342,  575 

integral,  111 

interdependent,  133,  145 

irrational.  111,  288,  313 

irreducible,  445 

linear,  112,  139 

literal.  111 

logarithmic,  390 

numerical.  111,  429,  459 

quadratic,  112,  298 

rational.  111 

reciprocal,  311,  438,  487 

roots  of.     See  Roots 

simple,  11:2,  118 

simultaneous  simple,  127,  143, 
508 

simultaneous,  of  higher  degree, 
135,  317,  514 

simultaneous  symmetric,  326 

solution  of,  112,  1.8,  483 

solution   of,    by  factorization, 
194,  309,  318 

transformation  of,  114, 129,436 
Errors  of  approximation,  55 
Evolution,  39,  56,  76,  83,  260,  276. 

490' 
Expectation,  value  of,  411 
Exponents,  integral,  39 

irrational,  376 

laws  of,  57,  279,  376 

rational,  279 
Expre.s.sions,  algebraic,  85 

finite  and  infinite,  85 

integral  and  fractional,  85 

rational  and  irrational,  86 

Factor,  14,  176 

highest  common,  196 


INDEX 


593 


Factor,  irreducible,  211 

prime,  177,  208,  212 

rationalizing,  285 
Factorial  n,  395 
Factorization,  178,  249 
Ferrari's  solution  of  biquadratic, 486 
Fractions,  32,  213 

continued,  566 

improper,  213 

irreducible,  37 

partial,  236 

proper,  213 

reciprocal,  219 

recurring,  continued,  572 
Functions,  88,  571,  585 

defined  by  power  series,  539 

expansion  of,  371,  548,  551 

integral,  85 

rational,  86 

symmetric,  245 
Fundamental  theorem  of  algebra, 
427,  588 

Graphs  of  e»i'nations,  139,  333 
of  numbers,  27,  38,  66 
of  variation  of  functions,  469 

Groups  of  things,  1 
equivalence  of,  1 
finite  and  infinite,  3 

Homogeneity,  87,  99 
Horner's  method,  453 
Hyperbola,  335 

Identities,  89 

Imaginaries,  conjugate,  295 
Incommensurable,  65 
Indeterminateness  of  rational  func- 
tions, cases  of,  223 
Induction,  mathematical,  424 
Inequalities,  solution  of,  340 
Inequality.     See  Equality 


Infinitesimal,  63 
Infinity  as  limit,  224,  229 
Interest,  compound,  390 
Interpolation,  371 
Inversions,  492 

Involution,  39,  56,  76,  82,  105,  276, 
489 

Lagrange's   formula   of   interpola- 
tion, 373 
Length,  26,  37,  66 
Limit  of  variable,  58 
Limits,    superior   and   inferior,    of 

assemblages,  582 
Logarithms,  39,^377 

characteristic  of,  381 

common,  379 

mantissa  of,  381 

modulus  of,  559 

natural,  558 

table  of,  384 

Maximum,  307,  467,  582 
Mean,  arithmetical,  355 

geometrical,  359 

harmonical,  362 
Measure,  26,  37,  65 
Minimum,  307,  467,  582 
Multinomial  theorem,  408 
Multiple,  lowest  common,  205 
Multiplication   of   integral   expres- 
sions, 97 

of  numbers,  14,20,35,52,72,489 

of  radicals,  275 

of  rational  expressions,  218 

of  series,  545 

Number,  cardinal,  2,  10 
complex,  71 
fractional,  33 
imaginary,  70 
integral,  18 
irrational,  46 


594 


A   COLLEGE    ALGEBRA 


Number,  natural,  6 

negative,  18 

positive,  18 

rational,  34 

real,  45 
Number  intervals,  579 

regions,  586 
Numbers,  theory  of,  211 

Odds,  410 

Ordinal,  7,  33,  45 

Ordinate,  138 

Origin,  137 

Oscillation  of  a  function. 


584 


Parabola,  333 

Parentheses,  rule  of,  95 
Part  (of  group),  3 
Permanences  of  sign,  446 
Permutations,  393 

odd  and  even,  492 
Polynomials  in  x,  87 

products  of,  103 
Powers.    See  Exponents  and  Invo- 
lution 

perfect,  260 
Power  series,  535 

convergence  of,  535 

products  of,  545 

quotients  of,  546 

reversion  of,  548 

transformation  of,  545 
Probability,  409 
Products,  continued,  252 

infinite,  564 
Progressions,  arithmetical,  354 

arithmetical,  of  higher  order, 
364 

geometrical,  357 

harmonical,  362 
Proportion,  347 

continued,  350 


Quadratics,  112,  298,  304 
simultaneous,  317 

Radical  expressions,  simple,  277 
Radicals,  271 

index  of,  271 

similar,  273 

simple,  271 
Radicand,  271 
Ratio,  69,  347 
Rationalization,  285 
Remainder  theorem,  169 
Resultants,  512 

properties  of,  514 
Rolle's  theorem,  467 
Roots  of  equations,  112,  426 

extraneous,  116 

imaginary,  444,  448 

infinite,  229,  306,  439 

irrational,  453 

location  of,  452,  458,  475 

multiple,  428,  463 

number  of,  427 

rational,  429 

superior  and  inferior  limits  of, 
430,  441,  466 

symmetric   functions   of,   305, 
434,  478 
Roots  of  integral  functions,  260 

cube,  266 

square,  262 
Roots  of  numbers,  39,  56,  76 

cube,  268,.  483 

principal,  271. 

square,  265,  292,  295,  296 

trigonometric    expression    of, 
490 

Scale,  complete,  6 

natural,  17 
Sequence  of  numbers,  58 

regular,  00 


INDEX 


o% 


Series,  alternating,  632 

binomial,  538,  553 

doubly  infinite,  543 

exponential,  537,  556 

geometric,  360 

hypergeometric,  529 

infinite,  520 

logarithmic,  537,  557 

recurring,  560 
Sign,  rules  of,  95 

Simultaneous,    127.      See    Equa- 
tions 
Sine,  488 

Solutions  of  systems  of  equations, 
128 

infinite,  230,  318 

integral,  342 

number  of,  517 
Square  root.     See  Roots 
Sturm's  theorem,  472 
Substitution,  principle  of,  128 
Subtraction  of  integral  expressions, 
93 

of  numbers,  16,  19,  35,  51,  72 

of  radicals,  274 


Subtraction    of    rational    expres- 
sions, 217 

of  series,  541 
Surds,^  291 
Symmetry,  absolute,  245 

cyclic,  248 

Taylor's  theorem,  461,  551 
Term,  86 

absolute,  426 
Transformation  of  equations,  114, 

129,  436 
Transposition  of  terms,  115 

Value,  absolute  or  numerical,  18, 

75,  488 
Variable,  58,  79 

continuous,  69 
Variation,  351 

of  integral  functions,  308,  469 
Variations  of  sign,  446 

Zero,  17 

as  limit,  63 

operations  with,  19,  25,  31 


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